Mathematics M.C.Q (1 Marks)

$ \mathrm{C}_1^2+\mathrm{C}_2^2+\mathrm{C}_3{ }^2=1 $

$ \overrightarrow{\mathrm{C}} \cdot(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})=|\mathrm{C}| \sqrt{9} \cos 60^{\circ} $

$ 2 \mathrm{C}_1+2 \mathrm{C}_2-\mathrm{C}_3=\frac{3}{2} $

$ \mathrm{C}_1-\mathrm{C}_3=1 $

$ \mathrm{C}_1+2 \mathrm{C}_2=\frac{1}{2} $

$ \mathrm{C}_1=\frac{\sqrt{2}}{3}+\frac{1}{2} $

$ \mathrm{C}_2=\frac{-1}{3 \sqrt{2}} $

$ \mathrm{C}_3=\frac{\sqrt{2}}{3}-\frac{1}{2}$

$ =\hat{i}-\hat{j}+\hat{k} $

$ (\vec{a} \times(\hat{i} \times \hat{j})) \times \hat{i}=\hat{k}+\hat{j} $

$ ((\vec{a} \times(\hat{i} \times \hat{j})) \times i) \times \hat{i}=\hat{j}-\hat{k} $

projection of $ \vec{a} \text { on } \hat{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} $

$ =\frac{1+1}{\sqrt{2}}=\sqrt{2}$

$ \overrightarrow{\mathrm{b}}=\mathrm{t}-2 \hat{\mathrm{j}}-2 \alpha \mathrm{t} \hat{\mathrm{k}} $

$ \text { so } \overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}}<0, \forall \mathrm{t} \in \mathrm{R} $

$ \alpha \mathrm{t}^2-12+6 \alpha \mathrm{t}<0 $

$ \alpha \mathrm{t}^2+6 \alpha \mathrm{t}-12<0, \forall \mathrm{t} \in \mathrm{R} $

$ \alpha<0 \text {, and } \mathrm{D}<0 $

$ 36 \alpha^2+48 \alpha<0 $

$ 12 \alpha(3 \alpha+4)<0 $

$ \frac{-4}{3}<\alpha<0 $

$ \text { also for } \mathrm{a}=0, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}<0 $

$ \text { hence a } \alpha \in\left(\frac{-4}{3}, 0\right]$

$ \vec{a} \cdot(\vec{c} \times \vec{b})-\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}$            $........(i)$

given $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$

$\Rightarrow(\vec{a} \times \vec{c}) \cdot \vec{b}=\vec{b} \cdot \vec{b}=|\vec{b}|^2=27 $    $.......(ii)$

$ \Rightarrow \vec{a} \cdot(\vec{c} \times \vec{b})=\left[\begin{array}{lll} \vec{a} & \vec{c} & \vec{b} \end{array}\right]=(\vec{a} \times \vec{c}) \cdot \vec{b}=27$

Now $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=3-6+3=0$   $........(iii)$

$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=3$           $....(iv) (given)$.

By $(i), (ii), (iii) \& (iv)$

$27-0-3=24$

Ratio $=\frac{8|(\vec{a} \times \vec{b})|}{|(\vec{a} \times \vec{b})|}=8$

$(\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=\overrightarrow{0}$

$\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}+\lambda \overrightarrow{\mathrm{b}}$

Now, $\overrightarrow{\mathrm{p}} . \overrightarrow{\mathrm{a}}=0$ (given)

So, $\vec{c} . \vec{a}+\lambda \vec{a} \cdot \vec{b}=0$

$(3-3-8)+\lambda(12+1-14)=0$

$\lambda=-8$

$\vec{p}=\vec{c}-8 \vec{b}$

$\vec{p}=-31 \hat{i}-11 \hat{j}-52 \hat{k}$

So, $\overrightarrow{\mathrm{p}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})$

$=-31+11+52 $

$=32$

$ (-4+7 d) \hat{i}-\hat{j}(-2+42)+\hat{k}(d-12) $

$ (7 d-4)^2+(40)^2+(d-12)^2=1800 $

$ 50 d^2-80 d-40=0 $

$ 5 d^2-8 d-4=0 $

$ 5 d^2-10 d-2 d-4 $

$ 5 d(d-2)+2(d-2)=0 $

$ d=2 \text { or } d=-\frac{2}{5} $

$ \because d>0, d=2 $

$ (a+1) \hat{i}+(b+2) \hat{j}+(c-7) \hat{k}=6 \hat{i}+2 \hat{j}-2 \hat{k} $

$ a+1=6 \& b+2=2, c-7=-2 $

$ a=5 \quad b=0 \quad c=5 $

$ |A B|=\sqrt{1+4+49}=\sqrt{54} $

$ |B C|=\sqrt{25+25}=\sqrt{50} $

$ |A C|=\sqrt{86+4+4}=\sqrt{44} $

Ans. 54

$ =|\overrightarrow{\mathrm{c}}|^2+4-0 $

$ \because \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} $

$ |\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}| $

$ 2=3|\overrightarrow{\mathrm{c}}| \sin \alpha $

$ |\overrightarrow{\mathrm{c}}|=\frac{2}{3} \operatorname{cosec} \alpha \quad \alpha \in\left[0, \frac{\pi}{3}\right] $

$ |\overrightarrow{\mathrm{c}}|_{\min }=\frac{2}{3} \times \frac{2}{\sqrt{3}} \quad \operatorname{cosec} \alpha \in\left[\frac{2}{\sqrt{3}}, \infty\right) $

$ \Rightarrow 27|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|_{\min }^2=27\left(\frac{16}{27}+4\right)=124$

$ \vec{a} \times(\vec{a} \times \vec{b})+\vec{a} \times(\vec{a} \times \vec{c})+\vec{a} \times(\vec{b} \times \vec{c}) $

$ =\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})$

$ (\vec{a} \cdot \vec{b}) \vec{a}-a^2 \vec{b}+(\vec{a} \cdot \vec{c}) \vec{a}-a^2 \vec{c}+(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k}) $

$ \Rightarrow-26 \vec{a}-29 \vec{b}+13 \vec{a}-29 \vec{c}+13 \vec{b}+26 \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})$

$ \Rightarrow \quad-13 \vec{a}-16 \vec{b}-3 \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k}) $

$ \Rightarrow \quad-13 \vec{a} \cdot \vec{b}-16 b^2-3 \vec{b} \cdot \vec{c}=\{\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})\} \cdot \vec{b}$

$\Rightarrow \quad(-13)(-26)-16(50)-3 \vec{b} \cdot \vec{c}=\left|\begin{array}{ccc}2 & -3 & 4 \\ 1 & 8 & 13 \\ 3 & 4 & -5\end{array}\right|$

$\Rightarrow \quad-462-3 \vec{b} \cdot \vec{c}=-396$

$\Rightarrow \quad \vec{b} \cdot \vec{c}=-22$

Hence $24-\vec{b} \cdot \vec{c}=46$

$ \Rightarrow(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\alpha-5,1,-2) $

$ \Rightarrow \mathrm{x}=\alpha-5, \mathrm{y}=1, \mathrm{z}=-2$           .............($1$)

$Image$

Area of $\Delta=5 \sqrt{6}$ (given)

$\frac{1}{2}|\vec{a} \times \vec{c}|=5 \sqrt{6}$

$\left\|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ x & 1 & -2\end{array}\right\|=10 \sqrt{6}$

$ \Rightarrow|-10 \hat{\mathrm{i}}-\hat{\mathrm{j}}(-2 \alpha-2 \mathrm{x})+\hat{\mathrm{k}}(\alpha-4 \mathrm{x})|=10 \sqrt{6} $

$ \Rightarrow(2 \alpha+2 \alpha-10)^2+(\alpha-4 \alpha+20)^2=500 $

$ \Rightarrow(4 \alpha-10)^2+(20-3 \alpha)^2=500 $

$ \Rightarrow 25 \alpha^2-80 \alpha-120 \alpha=0 $

$ \Rightarrow \alpha(25 \alpha-200)=0 $

$ \Rightarrow \alpha=8(\text { given } \alpha \text { is +ve number) } $

$ \Rightarrow \mathrm{x}=\alpha-5=3 $

$ |\overrightarrow{\mathrm{c}}|^2=\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2 $

$ =9+1+4 $

$ =14$

$ \mathrm{~A}=\frac{1}{2}|(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=\frac{\sqrt{21}}{2} $

$ \text { so, } \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+2 \hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\hat{\mathrm{i}}+\beta \hat{\mathrm{j}}$

$(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 2 \\ -1 & \beta & 0\end{array}\right|$

$ =\hat{\mathrm{i}}(-2 \beta)-\hat{\mathrm{j}}(2)+\hat{\mathrm{k}}(\beta+\alpha) $

$ |(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=\sqrt{4 \beta^2+4+(\alpha+\beta)^2}=\sqrt{21} $

$ 4 \beta^2+4+\alpha^2+\beta^2+2 \alpha \beta=21 $

$ \alpha^2+5 \beta^2-12=17 $

$ \alpha^2+5 \beta^2=29$

and $ \alpha \beta=-6$

and given $\alpha_i \beta$ are integers

so,

$\alpha=-3, \beta=2$

or

$ \alpha=3, \beta=-2 $

$ \left(\alpha_1, \beta_1\right)=(-3,2) $

$ \left(\alpha_2, \beta_2\right)=(3,-2) $

$ \alpha_1^2+\beta_1^2-\alpha_2 \beta_2=9+4+6=19$

$ \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})$

$ 1=\lambda(1+\mathrm{x}+5) $

$ 1=\lambda(\mathrm{x}+6) $

$ |\overrightarrow{\mathrm{d}}|=1 \quad \frac{1}{\lambda}=\mathrm{x}+6 $

$ |\lambda(\vec{b}+\vec{c})|=1 $

$ |\lambda((\mathrm{x}+2) \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})|=1 $

$ \lambda^2\left((\mathrm{x}+2)^2+6^2+2^2\right)=1 $

$ x^2+4 x+4+36+4=(x+6)^2 $

$ \mathrm{x}^2+4 \mathrm{x}+44=\mathrm{x}^2+12 \mathrm{x}+36 $

$ 8 \mathrm{x}=8, \mathrm{x}=1 $

$\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 4 & -5 \\ x & 2 & 3\end{array}\right|=(\vec{a} \times \vec{b}) \cdot \vec{c}$

$\left|\begin{array}{ccc}0 & 0 & 1 \\ -2 & 9 & -4 \\ x-2 & -1 & 3\end{array}\right|=2-9(x-2)$

$ =20-9 x $

$ \text { at } x=1 $

$ 20-9=11$

Option $4$ is correct

$\vec{b}=\hat{i}+2 \hat{j}-4 \hat{k}$

$(\vec{a} \times \vec{b}) \times \hat{i}=(\vec{a} \cdot \hat{i}) \vec{b}-(\vec{b} \cdot \hat{i}) \vec{a}$

$=-5 \vec{b}-\vec{a}$

$=(((-5 \vec{b}-\vec{a}) \times \hat{i}) \times \hat{i})$

$=((-11 \hat{j}+23 \hat{k}) \times \hat{i}) \times \hat{i}$

$\Rightarrow(11 \hat{k}+23 \hat{j}) \times \hat{i}$

$\Rightarrow(11 \hat{j}-23 \hat{k})$

$\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})=11-23=-12$

$ |\overrightarrow{c}-\overrightarrow{a}|=2 \sqrt{2} $

$ |\mathrm{c}|^2+|\mathrm{a}|^2-2 \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=8 $

$ |z|^2+38-12|z|=8 $

$ |z|^2-12|z|+30=0 $

$ |z|=\frac{12 \pm \sqrt{144-120}}{2} $

$ =\frac{12 \pm 2 \sqrt{6}}{2} $

$ |z|=6+\sqrt{6} $

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{\ell} & \hat{j} & \hat{k} \\ 6 & 1 & -1 \\ 1 & 1 & 0\end{array}\right|$

$ \hat{\ell}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}} $

$ |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{27} $

$ |(\overrightarrow{\mathrm{a}} \times \mathrm{b}) \times \mathrm{z}|=\sqrt{27}(6+\sqrt{6}) \frac{\sqrt{3}}{2} $

$ \frac{9}{2}(6+\sqrt{6})$

Orthocentre and centroid will be same

$\mathrm{G}\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)$

 Mid-point of $ \mathrm{AB} \text { is } \mathrm{D}\left(\frac{3}{2}, 2, \frac{3}{2}\right) $

$ \therefore \ell_1=\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}} $

$ \ell_1=\sqrt{\frac{1}{6}}=\ell_2=\ell_3 $

$ \therefore \ell_1^2+\ell_2^2+\ell_3^2=\frac{1}{2}$

$\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$

Dot product with $\overrightarrow{\mathrm{a}}$ on both sides

$\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=-6$

Dot product with $\vec{b}$ on both sides

$ \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=-48 $

$\overrightarrow{\mathrm{c}} \overrightarrow{\mathrm{c}}=4|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^2+9|\overrightarrow{\mathrm{b}}|^2 $

$ |\overrightarrow{\mathrm{c}}|^2=4\left[|\mathrm{a}|^2|\mathrm{~b}|^2-(\mathrm{a} \cdot \overrightarrow{\mathrm{b}})^2\right]+9|\overrightarrow{\mathrm{b}}|^2 $

$ |\overrightarrow{\mathrm{c}}|^2=4\left[(1)(4)^2-(4)\right]+9(16) $

$ |\overrightarrow{\mathrm{c}}|^2=4[12]+144 $

$ |\overrightarrow{\mathrm{c}}|^2=48+144 $

$ |\overrightarrow{\mathrm{c}}|^2=192 $

$ \therefore \cos \theta=\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}|} $

$ \therefore \cos \theta=\frac{-48}{\sqrt{192} \cdot 4} $

$ \therefore \cos \theta=\frac{-48}{8 \sqrt{3} \cdot 4} $

$ \therefore \cos \theta=\frac{-3}{2 \sqrt{3}} $

$ \therefore \cos \theta=\frac{-\sqrt{3}}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$

$|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=18$

$|\vec{a}|^2=6$

Also $1+\alpha^2+\beta^2=6$

$\left(\alpha^2+\beta^2\right)|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta$
$=(5)(6)(6)\left(\frac{1}{2}\right)$

$=(5)(6)(6)\left(\frac{1}{2}\right)$

$=90$

$\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z\end{array}\right|=14 \hat{i}+10 \hat{j}-20 \hat{k}$

$\begin{aligned} & \Rightarrow \hat{i}(z-4 y)-\hat{j}(5 z-4 x)+\hat{k}(5 y-x)=14 \hat{i}+10 \hat{j}-20 \hat{k} \\ & z-4 y=14,4 x-5 z=10,5 y-x=-20 \\ & (a-b+i) \cdot \vec{c}=-3 \\ & (2 \hat{i}+3 \hat{j}-2 \hat{k}) \cdot \vec{c}=-3 \\ & 2 x+3 y-2 z=-3 \\ & \therefore x=5, y=-3, z=2 \\ & |\vec{c}|^2=25+9+4=38\end{aligned}$

$ \vec{b}=-\hat{i}-8 \hat{j}+2 \hat{k} $

$ \overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}+\mathrm{c}_2 \hat{\mathrm{j}}+\mathrm{c}_3 \mathrm{k} $

$ \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} $

$ (\vec{b}-\vec{c}) \times \vec{a}=0 $

$ \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=\lambda \vec{\alpha} $

$ \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}+\lambda \vec{\alpha} $

$ -\hat{i}-8 \hat{j}+2 k=\left(4 \hat{i}+c_2 \hat{j}+c_3 k\right)+\lambda(\hat{i}+\hat{j}+k) $

$ \lambda+4=-1 \Rightarrow \lambda=-5 $

$ \lambda+\mathrm{c}_2=-8 \Rightarrow \mathrm{c}_2=-3 $

$ \lambda+c_3=2 \Rightarrow c_3=7$

$ \overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \mathrm{k} $

$ \cos \theta=\frac{12-12+7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{2 \sqrt{481}} $

$ \tan ^2 \theta=\frac{625 \times 3}{49} $

$ {\left[\tan ^2 \theta\right]=38} $

$ (2 \overrightarrow{\mathrm{b}}+4 \overrightarrow{\mathrm{a}}) \times \overrightarrow{\mathrm{c}}=0 $

$ \overrightarrow{\mathrm{c}}=\lambda(4 \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}})=\lambda(8 \hat{\mathrm{i}}-14 \hat{\mathrm{j}}+30 \hat{\mathrm{k}}) $

$ \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=130 $

$ 8 \lambda+42 \lambda+210 \lambda=130 $

$ \lambda=\frac{1}{2} $

$ \overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+15 \hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=8+7+15=30$

$\overrightarrow{\mathrm{v}} $$ =\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\hat{\mathrm{i}} $

$ =5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

and $\overrightarrow{\mathrm{c}}+\hat{\mathrm{i}}=\overrightarrow{\mathrm{p}}$

So,

$ \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{p}} $

$ \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{0} $

$ \overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}})=\overrightarrow{0} $

$ \Rightarrow \overrightarrow{\mathrm{p}}=\lambda(\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}}) $

$ \overrightarrow{\mathrm{c}}+\mathrm{i}=\lambda(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) $

$ \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{a}} \cdot \hat{\mathrm{i}}=\lambda \overline{\mathrm{a}} \cdot(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) $

$ -29+2=\lambda(14+40) $

$ \lambda=-\frac{1}{2}$

$ \overrightarrow{\mathrm{c}} \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\hat{\mathrm{i}} \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=\lambda(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) $

$ \quad=-\frac{1}{2}(-14+8)+2=5$

$ \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{c}}=17 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=20 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-12 \hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=-14 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}-14 \hat{\mathrm{k}} $

$ (\overrightarrow{\mathrm{r}}-(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})) \times \overrightarrow{\mathrm{a}}=0 $

$ \mathrm{r}-(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\lambda \overrightarrow{\mathrm{a}} $

$ \overrightarrow{\mathrm{r}}=\lambda \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} $

$ B \mathrm{r} \overrightarrow{\mathrm{r}} \cdot(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})=0 $

$ \Rightarrow(\lambda \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})=0 $

$ \Rightarrow \lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}-\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=0 $

$ \lambda=\frac{\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}}{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}}=\frac{294-227}{-389=204}=\frac{-67}{593} $

$ \therefore \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}-\frac{67}{593} \overrightarrow{\mathrm{a}} $

$ \Rightarrow 593 \overrightarrow{\mathrm{r}}+67 \overrightarrow{\mathrm{a}}=593(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) $

$ \Rightarrow|\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^2=569$

$ (\vec{a}+\vec{b}) \times \vec{c}+(-2 \vec{a}+3 \vec{b}) \times \vec{c}=0 $

$ \Rightarrow(\vec{a}+\vec{b})-2 \vec{a}+3 \vec{b}) \times \vec{c}=0 $

$ \Rightarrow \vec{c}=\lambda(4 \vec{b}-\vec{a}) $

$ \Rightarrow=\lambda(44 \hat{i}-4 \hat{j}+4 \hat{k}-4 \hat{i}+\hat{j}-\hat{k}) $

$ =\lambda(40 \hat{i}-3 \hat{j}+3 \hat{k})$

Now

$ (8 \hat{i}-2 \hat{j}+2 \hat{k}+33 \hat{i}-3 \hat{j}+3 \hat{k}) \cdot \lambda(40 \hat{i}-3 \hat{j}+3 \hat{k})=1670 $

$ \Rightarrow(41 \hat{i}-5 \hat{j}+5 \hat{k}) \cdot(40 \hat{i}-3 \hat{j}+3 \hat{k}) \times \lambda=1670) $

$ \Rightarrow(1640+15+15) \lambda=1670 \Rightarrow \lambda=1 $

$ \text { so } \overrightarrow{\mathrm{c}}=40 \hat{i}-3 \hat{j}-3 \hat{k} $

$ \Rightarrow|\overrightarrow{\mathrm{c}}|^2=1600+9+9=1618$

$ \bar{a}=2 \hat{i}+\hat{j}-\hat{k} $

$ \bar{a} \times \bar{r}=\bar{a} \times \bar{b} ; \bar{a} \cdot \bar{r}=0 $

$ \Rightarrow \bar{a} \times(\bar{r}-\bar{b})=\overline{0} $

$ \Rightarrow \bar{a}=\lambda(\bar{r}-\bar{b}) $

$ \bar{a} \cdot \bar{a}=\lambda(\bar{a} \cdot \bar{r}-\bar{a} \cdot \bar{b}) $

$ 14=-7 \lambda \Rightarrow \lambda=-2 $

$ \frac{-\bar{a}}{2}=\bar{r}-\bar{b} \Rightarrow \bar{r}=\bar{b}-\frac{\bar{a}}{2} $

$ =\frac{2 \bar{b}-\bar{a}}{2}=\frac{3 \hat{i}+\hat{k}}{2}$

Statement ($I$) is incorrect

$ \cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{c} \geq-\frac{3}{2} $

$ 2 \mathrm{~A}+2 \mathrm{~B}+2 \mathrm{C}=2 \pi $

$ \cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C} $

$ =-1-4 \cos \mathrm{A} \cdot \cos \mathrm{B} \cdot \cos \mathrm{C} $

$ \geq-1-4 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} $

$ =-\frac{3}{2}$

Statement ($II$) is correct.

$\overline{ PG }=\frac{\overline{ a }+\overline{ b }+\overline{ c }}{3}$

$\Rightarrow \overline{ a }+\overline{ b }+\overline{ c }=3 \overline{ PG }=\overline{ PQ }$

$\overrightarrow{ b } \cdot \overrightarrow{ c }=\overrightarrow{ b } \cdot(2 \overrightarrow{ a } \times \overrightarrow{ b })-3 \overrightarrow{ b } \cdot \overrightarrow{ b }$

$=-3| b |^2$

$=-48$

$|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2$

$|-3 \vec{a} \times \vec{b}+4 \vec{b} \times \vec{a}|^2$

$|-3 \vec{a} \times \vec{b}-4 \vec{a} \times \vec{b}|^2$

$|-7 \vec{a} \times \vec{b}|^2$

$\left(-7|\vec{a}| \times|\vec{b}| \sin \left(\frac{\pi}{4}\right)\right)^2$

$49 \times 4 \times 9 \times \frac{1}{2}=882$

$A, B, C, D$ are coplanar points, then

$\begin{array}{l}\Rightarrow\left|\begin{array}{ccc}1-3 & 2+4 & -1-2 \\ -2-3 & -1+4 & 3-2 \\ 5-3 & -2 \alpha+4 & 4-2\end{array}\right|=0\end{array}$

$\Rightarrow \alpha=\frac{73}{17}$

$\begin{aligned} & \left|\begin{array}{ccc}1 & -1 & 1 \\ \alpha & 2 & -1 \\ 8 & \alpha-6 & 3\end{array}\right|=0 \\ & \Rightarrow \alpha^2-2 \alpha-8=0 \\ & \Rightarrow(\alpha-4)(\alpha+2)=0 \\ & \therefore \alpha=4,-2\end{aligned}$

So, $[\overrightarrow{ b }-\overrightarrow{ a } \overrightarrow{ c }-\overrightarrow{ a } \overrightarrow{ d }-\overrightarrow{ a }]=0$

$(\overrightarrow{ b }-\overrightarrow{ a }) \cdot((\overrightarrow{ c }-\overrightarrow{ a }) \times(\overrightarrow{ d }-\overrightarrow{ a }))=0$

${[\overrightarrow{ b } \overrightarrow{ c } \overrightarrow{ d }]-[\overrightarrow{ b } \overrightarrow{ c } \overrightarrow{ a }]-[\overrightarrow{ b } \overrightarrow{ a } \overrightarrow{ d }]-[\overrightarrow{ a } \overrightarrow{ c } \overrightarrow{ d }]=0}$

$\Rightarrow[\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }]=[\overrightarrow{ c } \overrightarrow{ d } \overrightarrow{ b }]+[\overrightarrow{ b } \overrightarrow{ d } \overrightarrow{ a }]+[\overrightarrow{ d } \overrightarrow{ c } \overrightarrow{ a }]$

$C _2 \rightarrow C _2- C _1, C _3 \rightarrow C _3- C _1$

$\left|\begin{array}{lll}a & 1-a & 1-a \\ 1 & b -1 & 0 \\ 1 & 0 & c -1\end{array}\right|=0$

$a(b-1)(c-1)-(1-a)(c-1)+(1-a)(1-b)=0$

$a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)=0$

$\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$

$\Rightarrow-1+\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$

$\Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$

$\overrightarrow{ a } \cdot \overrightarrow{ c }=\frac{1}{2}, \overrightarrow{ a } \cdot \overrightarrow{ b }=\frac{1}{2}$

$\therefore \overrightarrow{ b } \cdot \overrightarrow{ d }=\frac{1}{2}$

$(\overrightarrow{ a } \times \overrightarrow{ b }) \cdot(\overrightarrow{ c } \times \overrightarrow{ d })=\overrightarrow{ a } \cdot(\overrightarrow{ b } \times(\overrightarrow{ c } \times \overrightarrow{ d }))$

$=\overrightarrow{ a } \cdot((\overrightarrow{ b } \cdot \overrightarrow{ d }) \overrightarrow{ c }-(\overrightarrow{ b } \cdot \overrightarrow{ c }) \overrightarrow{ d })$

$=(\overrightarrow{ a } \cdot \overrightarrow{ c })(\overrightarrow{ b } \cdot \overrightarrow{ d })=\frac{1}{4}$

Taking cross product with $\overrightarrow{ a }$

$\Rightarrow \quad \vec{a} \times(\vec{a} \times \vec{b})=\vec{a} \times(\hat{i}-\hat{j})$

$\Rightarrow \quad(\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}=\hat{i}+\hat{j}+2 \hat{k}$

$\Rightarrow \quad \vec{a}-3 \vec{b}=\hat{i}+\hat{j}+2 \hat{k}$

$\Rightarrow \quad 2 \vec{a}-6 \vec{b}=2 \hat{i}+2 \hat{j}+4 \hat{k}$

$\Rightarrow \quad \quad \quad \vec{a}-6 \vec{b}=3 \hat{i}+3 \hat{j}+3 \hat{k}$

$\overrightarrow{ b }=\hat{ i }+\hat{ k }$

$\overrightarrow{ c }=\hat{ i }+2 \hat{ j }-3 \hat{ k }$

$\overrightarrow{ r } \times \overrightarrow{ a }=\overrightarrow{ c } \times \overrightarrow{ a } \Rightarrow(\overrightarrow{ r }-\overrightarrow{ c }) \times \overrightarrow{ a }=0$

$\therefore \overrightarrow{ r }=\overrightarrow{ c }+\lambda \overrightarrow{ a }$

$\overrightarrow{ r } \cdot \overrightarrow{ b }=0 \Rightarrow \overrightarrow{ c } \cdot \overrightarrow{ b }+\lambda \quad \overrightarrow{ b } \cdot \overrightarrow{ a }=0$

$-2+\lambda(7)=0 \Rightarrow \lambda=\frac{2}{7}$

$\therefore \overrightarrow{ r }=\overrightarrow{ c }+\frac{2 \overrightarrow{ a }}{7}=\frac{1}{7}(11 \hat{ i }-11 \hat{ k })$

$|\overrightarrow{ r }|=\frac{11 \sqrt{2}}{7}$

$|\vec{a} \times \vec{b}|^2+|\vec{a} \cdot \vec{b}|^2=|\vec{a}|^2 \times|\vec{b}|^2$

$\Rightarrow(\vec{a} \cdot \vec{b})^2=84-48=36$

Let $\left|a_1\right| \leq\left|a_2\right| \leq\left|a_3\right|$

$|\vec{a}|^2=\left|a_1\right|^2+\left|a_2\right|^2+\left|a_3\right|^2 \geq\left(a_3\right)^2$

$|\vec{a}| \geq\left|a_3\right|=\max \left\{\left|a_1\right|,\left|a_2\right|,\left|a_3\right|\right\}$

$A$ is true

$|\vec{a}|^2=\left| a _1\right|^2+\left| a _2\right|^2+\left| a _3\right|^2 \leq\left| a _3\right|^2+\left| a _3\right|^2+\left| a _3\right|^2$

$|\overrightarrow{ a }|^2 \leq 3\left| a _3\right|^2$

$|\overrightarrow{ a }| \leq \sqrt{3}\left| a _3\right|=\sqrt{3} \max \left\{\left| a _1\right|,\left| a _2\right|,\left| a _3\right|\right\}$

$\leq 3 \max \left\{\left| a _1\right|,\left| a _2\right|,\left| a _3\right|\right\}$

$(2)$ is true

$(\overrightarrow{ b }-\overrightarrow{ a })-(\overrightarrow{ c }-\overrightarrow{ b })+(\overrightarrow{ d }-\overrightarrow{ a })-(\overrightarrow{ c }-\overrightarrow{ d })= k \overline{ FE }$

$2(\overrightarrow{ b }+\overrightarrow{ d })-2(\overrightarrow{ a }-\overrightarrow{ c })= k \overrightarrow{ FE }$

$2(2 \overrightarrow{ f })-2(2 \overrightarrow{ e })= k \overline{ FE }$

$4(\overrightarrow{ f }-\overrightarrow{ e })= k \overline{ FE }$

$-4 \overline{ FE }= k \overrightarrow{ FE }$

$k=-4$

$\Rightarrow \overrightarrow{ b }=\lambda(-2 \hat{ i }-2 \hat{ j }+2 \hat{ k })$

$|\overrightarrow{ b }|=|\overrightarrow{ a }| \quad \therefore \sqrt{6}=\sqrt{12}|\lambda| \Rightarrow \lambda=\pm \frac{1}{\sqrt{2}}$

$\left(\lambda=\frac{1}{\sqrt{2}} \text { rejected } \because \overrightarrow{ b } \text { makes acute angle with y axis }\right)$

$\overrightarrow{ b }=-\sqrt{2}(-\hat{ i }-\hat{ j }+\hat{ k })$

$\frac{(3 \overrightarrow{ a }+\sqrt{2} \overrightarrow{ b }) \cdot \overrightarrow{ c }}{|\overrightarrow{ c }|}=3 \sqrt{2}$

Let $\quad \vec{c}=x \hat{i}+y \hat{j}+z \hat{k}$

$\Rightarrow 15 x-20 y-25 z+25=0$

$\Rightarrow 3 x-4 y-5 z=-5$

Also $x+y+z=4$

$\text { and } \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|}=1 \Rightarrow 4 x+3 y=5$

$\Rightarrow \quad \vec{c}=2 \hat{i}-\hat{j}+3 \hat{k}$

Projection of $\overrightarrow{ c }$ or $\overrightarrow{ b }=\frac{25}{5 \sqrt{2}}=\frac{5}{\sqrt{2}}$

$2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a}=2 \vec{a} \cdot \vec{b}-2 \vec{b} \cdot \vec{c}-2 \vec{c} \cdot \vec{a}$

$4 \vec{a} \cdot \vec{c}=0$

$B$ is incorrect

$|\overrightarrow{ a }+\lambda \overrightarrow{ c }|^2 \geq|\overrightarrow{ a }|^2$

$\lambda^2 c ^2 \geq 0$

True $\forall \lambda \in R$ $(A)$ is correct.

$\overrightarrow{ b }=\hat{ i }-\hat{ j }+2 \hat{ k }$

$\overrightarrow{ c }=\hat{5 i }-3 \hat{ j }+3 \hat{ k }$

$(\overrightarrow{ r }-\overrightarrow{ c }) \times \overrightarrow{ b }=0, \overrightarrow{ r } \cdot \overrightarrow{ a }=0$

$\Rightarrow \overrightarrow{ r }-\overrightarrow{ c }=\lambda \overrightarrow{ b }$

$\text { Also, }(\overrightarrow{ c }+\lambda \overrightarrow{ b }) \cdot \overrightarrow{ a }=0$

$\Rightarrow \overrightarrow{ a } \cdot \overrightarrow{ c }+\lambda(\overrightarrow{ a } \cdot \overrightarrow{ b })=0$

$\therefore \lambda=\frac{\overrightarrow{ a } \cdot \overrightarrow{ c }}{\overrightarrow{ a } \cdot \overrightarrow{ b }}=\frac{-8}{5}$

$\overrightarrow{ r }=\frac{5(5 \hat{ i }-3 \hat{ i }+3 \hat{ k })-8(\hat{ i }-\hat{ j }+2 \hat{ k })}{5}$

$\overrightarrow{ r }=\frac{17 \hat{ i }-7 \hat{ j }+\hat{ k }}{5}$

$|\overrightarrow{ r }|^2=\frac{1}{25}(289+50)$

$25|\overrightarrow{ r }|^2=339$

$\vec{b}=\hat{i}-3 \hat{j}+5 \hat{k}$

$\vec{a} \cdot \hat{b}=\frac{5-3-15}{\sqrt{35}}=-\frac{-13}{\sqrt{35}}$

$=(2 \hat{i}+4 \hat{j}-2 \hat{k})-(-2 \hat{i}+\hat{j}-3 \hat{k})$

$=4 \hat{i}+3 \hat{j}+\hat{k}$

$\overline{A C}=\overline{O C}-\overline{O A}=-2 \hat{i}+\hat{j}+2 \hat{k}$

$\overline{A B} \times \overrightarrow{A C}=5 \hat{i}-10 \hat{j}+10 \hat{k}$

$\overline{O P}=-\hat{i}-2 \hat{j}+3 \hat{k}$

Projection

$=\frac{(\overline{O P}) \cdot(\overline{A B} \times \overline{A C})}{|\overrightarrow{A B} \times \overrightarrow{A C}|}=3$

$\alpha= 2 , \beta= 4 , \gamma-\delta= 3$

$\frac{1}{2}|\overline{ AB } \times \overline{ AC }|=5 \sqrt{6}$

$(\delta-9)^2+(2 \delta+12)^2+100=600$

$\Rightarrow \delta=5, \gamma=8$

Hence $\overline{ CB } \cdot \overline{ CA }=60$

Area of $\triangle ABC$ is $\frac{1}{2}|\overrightarrow{ AB } \times \overrightarrow{ AC }|$ $\overrightarrow{ AB }=\frac{\overrightarrow{ r }-2 \overrightarrow{ q }}{3}, \overrightarrow{ AC }=\frac{-\overrightarrow{ r }-\overrightarrow{ q }}{3}$

Area of $\triangle ABC =\frac{1}{6}|\vec{q} \times \overrightarrow{ r }|$

$\frac{\operatorname{Area}(\triangle PQR )}{\operatorname{Area}(\triangle ABC )}=3$

$\overrightarrow{ a } \times \overrightarrow{ c }-\overrightarrow{ b } \times \overrightarrow{ c }=\overrightarrow{0}$

$(\overrightarrow{ a }-\overrightarrow{ b }) \times \overrightarrow{ c }=0 \Rightarrow(\overrightarrow{ a }-\overrightarrow{ b }) \text { is paralleled to } \overrightarrow{ c }$

$\overrightarrow{ a }-\overrightarrow{ b }=\mu \overrightarrow{ c } \text {, where } \mu \text { is a scalar }$

$-2 \hat{ i }+7 \hat{ j }+2 \lambda \hat{ k }=\mu \cdot \overrightarrow{ c }$

Now $\vec{a} \cdot \overrightarrow{ c }=7$ gives $2 \lambda^2+12=7 \mu$

And $\vec{b} \cdot \vec{c}=-\frac{43}{2}$ gives $4 \lambda^2+82=43 \mu$

$\mu=2$ and $\lambda^2=1$

$|\vec{a} \cdot \vec{b}|=8$

$\Rightarrow(\overrightarrow{ r }-\overrightarrow{ c }) \times \overrightarrow{ b }=0$

$\Rightarrow \overrightarrow{ r }-\overrightarrow{ c }=\lambda \overrightarrow{ b }$

$\Rightarrow \overrightarrow{ r }=\overrightarrow{ c }+\lambda \overrightarrow{ b }$

And given that $\overrightarrow{ r } \cdot \overrightarrow{ a }=0$

$\Rightarrow(\overrightarrow{ c }+\lambda \overrightarrow{ b }) \cdot \overrightarrow{ a }=0$

$\Rightarrow \overrightarrow{ c } \cdot \overrightarrow{ a }+\lambda \overrightarrow{ b } \cdot \overrightarrow{ a }=0$

$\Rightarrow \lambda=\frac{-\overrightarrow{ c } \cdot \overrightarrow{ a }}{\overrightarrow{ b } \cdot \overrightarrow{ a }}$

Now $\overrightarrow{ r } \cdot \overrightarrow{ c }=(\overrightarrow{ c }+\lambda \overrightarrow{ b }) \cdot \overrightarrow{ c }$

$=\left(\overrightarrow{ c }-\frac{\overrightarrow{ c } \cdot \overrightarrow{ a }}{\overrightarrow{ b } \cdot \overrightarrow{ a }}\right) \cdot \overrightarrow{ c }$

$=|\overrightarrow{ c }|-\left(\frac{\overrightarrow{ c } \cdot \overrightarrow{ a }}{\overrightarrow{ b } \cdot \overrightarrow{ a }}\right)(\overrightarrow{ b } \cdot \overrightarrow{ c })$

$=74-\left[\frac{15}{3}\right] 8$

$=74-40=34$

$\overrightarrow{ b }=\hat{ i }-\lambda \hat{ j }+2 \hat{ k }$ $\Rightarrow(\overrightarrow{ b }-\overrightarrow{ a }) \times((\overrightarrow{ a }+\overrightarrow{ b }) \times(\overrightarrow{ a } \times \overrightarrow{ b }))=8 \hat{ i }-40 \hat{ j }-24 \hat{ k }$

$\Rightarrow((\overrightarrow{ a }-\overrightarrow{ b }) \cdot(\overrightarrow{ a }+\overrightarrow{ b }))(\overrightarrow{ a } \times \overrightarrow{ b })=8 \hat{ i }-40 j -24 \hat{ k }$

$\Rightarrow 8(\overrightarrow{ a } \times \overrightarrow{ b })=8 \hat{ i }-40 \hat{ j }-24 \hat{ k }$

$\text { Now, } \overrightarrow{ a } \times \overrightarrow{ b }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2\end{array}\right|$

$=(4-3 \lambda) \hat{ i }-(2 \lambda+3) \hat{ j }+\left(-\lambda^2-2\right) \hat{ k }$

$\Rightarrow \lambda=1$

$\therefore \overrightarrow{ a }=\hat{ i }+2 \hat{ j }-3 \hat{ k }$

$\overrightarrow{ b }=\hat{ i }-\hat{ j }+2 \hat{ k }$

$\Rightarrow \lambda=1$

$\therefore \overrightarrow{ a }=\hat{ i }+2 \hat{ j }-3 \hat{ k }$

$\overrightarrow{ b }=\hat{ i }-\hat{ j }+2 \hat{ k }$

$\Rightarrow \overrightarrow{ a }+\overrightarrow{ b }=2 \hat{ i }+\hat{ j }-\hat{ k }, \overrightarrow{ a }-\overrightarrow{ b }=3 \hat{ j }-5 \hat{ k }$ $\Rightarrow(\overrightarrow{ a }+\overrightarrow{ b }) \times(\overrightarrow{ a }-\overrightarrow{ b })=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 1 & -1 \\ 0 & 3 & -5\end{array}\right|=2$

$\hat{ i }+10 \hat{ j }+6 \hat{ k }$

$\therefore \text { required answer }=4+100+36=140$

$\vec{a} \times(2 \vec{b}+3 \vec{c})=0$

$\vec{a}=\lambda(2 \vec{b}+3 \vec{c})$

$|\vec{a}|^2=\lambda^2|2 \vec{b}+3 \vec{c}|^2$

$|\vec{a}|^2=\lambda^2\left(4|\vec{b}|^2+9|\vec{c}|^2+12 \vec{b} \cdot \vec{c}\right)$

$31=31 \lambda^2 \Rightarrow \lambda=\pm 1$

$\vec{a}=\pm(2 \vec{b}+3 \vec{c})$

$\frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{2|\vec{b} \times \vec{c}|}{2 \vec{b} \cdot \vec{b}+3 \vec{c} \cdot \vec{b}}$

$|\vec{b} \times \vec{c}|^2=|\vec{b}|^2|\vec{c}|^2-(\vec{b} . \vec{c})^2=\frac{3}{4}$

$\frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|}=\frac{2 \times \frac{\sqrt{3}}{2}}{2 \cdot \frac{1}{4}-\frac{3}{2}}=-\sqrt{3}$

$\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\right)^2=3$