Mathematics M.C.Q (1 Marks)

$\text { Area }=\frac{1}{2}|\overrightarrow{B D} \times \overline{A C}|=18$

$\overrightarrow{A C} \times \overline{B D}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & -3 & -3 \\ 8 & 5 & -\lambda \end{array}\right|$

$=(3 \lambda+15) \hat{i}-\hat{j}(-24)+\hat{k}(-24)$

$\overline{A C} \times \overline{B D}=(3 \lambda+15) \hat{i}+24 \hat{j}-24 \hat{k}$

$=\sqrt{(3 \lambda+15)^2+(24)^2+(24)^2}=36 $

$ =\lambda^2+10 \lambda+9=0 $

$=\lambda=-1,-9$

$|\lambda| \leq 5 \Rightarrow \lambda=-1$

$5-6 \lambda=5-6(-1)=11$

$\overrightarrow{ b } \times \overrightarrow{ c }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & -2 & -2 \\ -1 & 4 & 3\end{array}\right|=2 \hat{i}-\hat{ j }+2 \hat{ k }$

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

$\vec{a} \cdot \vec{d}=18$

$\lambda=2$

So $\overrightarrow{ d }=2(2 \hat{ i }-\hat{ j }+2 \hat{ k })$

$\overrightarrow{ d } \times \overrightarrow{ a }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 4 & -2 & 4 \\ 2 & 3 & 4\end{array}\right|=-20 \hat{ i }-8 \hat{ j }+16 \hat{ k }$

$|\overrightarrow{ d } \times \overrightarrow{ a }|^2=720$

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

$\vec{a} \|^{ I }(\overrightarrow{ c }-\overrightarrow{ b })$

$\therefore \overrightarrow{ a }=\lambda(\overrightarrow{ c }-\overrightarrow{ b })$

$(6,9,12)=\lambda[x-\alpha, y-11, z+2]$

$\frac{x-\alpha}{2}=\frac{y-11}{3}=\frac{z+2}{4}$

$4 y-44=3 z+6$

$4 y-3 z=50$

$6 x+9 y+12 z=-12$

$2 x+3 y+4 z=-4$

$(\because x-2 y+z=5)$

$2 x-4 y+2 z=10$

$+\quad-$

$7 y+2 z=-14 \ldots(2)$

$8 y-6 z=100$

$21 y+6 z=-42$

$29 y =58$

$y =2, z =-14$

$\therefore x -4-14=5$

$x =23$

$\overline{ c }=(23,2,-14)$

$\overline{ c } \cdot(1,1,1)=23+2-14=11$

$=\frac{1}{2} \overrightarrow{ AB } \times \overrightarrow{ AC }+\frac{1}{2} \overrightarrow{ AC } \times \overrightarrow{ AD }$

$=\frac{1}{2}(\overrightarrow{ AB }-\overrightarrow{ AD }) \times \overrightarrow{ AC }$

$=\frac{1}{2}(8 \hat{ j }+12 \hat{ k }) \times(-4)(\hat{ i }+\hat{ j }-\hat{ k })$

$=\frac{1}{2}\left|\begin{array}{lll}\hat{ i } & \hat{ j } & \hat{ k } \\ 0 & 8 & 12 \\ 1 & 1 & -1\end{array}\right|$

$=(-2)(-20 \hat{ i }+12 \hat{ j }-8 \hat{ k })$

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

$\therefore \text { Area }=|\overrightarrow{ v }|=8 \sqrt{25+9+4}=8 \sqrt{38} \text { Ans. }$

$\left(\begin{array}{l}\overrightarrow{ AB }=-\hat{ i }+\hat{ j }+4 \hat{ k } \\ \overrightarrow{ AD }=-\hat{ i }-7 \hat{ j }-8 \hat{ k } \\ \overrightarrow{ AC }=-4 \hat{ i }-4 \hat{ j }+4 \hat{ k }\end{array}\right)$

$\overrightarrow{ n }_1=\left|\begin{array}{lll}\hat{ i } & \hat{ j } & \hat{ j } \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right|=\hat{ i }-\hat{ j }-\hat{ k }$

$\overrightarrow{ n }_2=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ j } \\ 1 & -1 & 0 \\ 1 & 0 & -1\end{array}\right|=\hat{i}+\hat{j}+\hat{ k }$

$\overrightarrow{ a }=\lambda\left|\overrightarrow{ n }_2 \times \overrightarrow{ n }_2\right|$

$=\lambda\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & -1 & -1 \\ 1 & 1 & 1\end{array}\right|=\lambda(-2 \hat{ j }+2 \hat{ k })$

$\overrightarrow{ a } \cdot \overrightarrow{ b }=\lambda|0+4+2|=6$

$\Rightarrow \lambda=1$

$\vec{\alpha}=-2 \hat{ j }+2 \hat{ k }$

$\cos \theta=\frac{\overrightarrow{ a } \cdot \overrightarrow{ b }}{| a || b |}$

$\cos \theta=\frac{6}{2 \sqrt{2} \times 3}=\frac{1}{\sqrt{2}}$

$\theta=\frac{\pi}{4}$

$\text { Now } \quad|\overrightarrow{ a } \cdot \overrightarrow{ b }|^2+|\overrightarrow{ a } \times \overrightarrow{ b }|^2=| a |^2| b |^2$

$36+\left|\overrightarrow{ a } \times b ^2\right|=8 \times 9=72$

$|\vec{a} \times b|^2=36$

$|\vec{a} \times \vec{b}|=6$

$\vec{b} \cdot(\vec{a} \times \vec{c})=27, \vec{a} \cdot \vec{b}=0$

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

Let $\theta$ be angle between $\vec{b}, \vec{a} \times \vec{c}$

Then $|\vec{b}| \cdot|\vec{a} \times \vec{c}| \sin \theta=3 \sqrt{14}$

$|\vec{b}| \cdot|\vec{a} \times \vec{c}| \cos \theta=27$

$\Rightarrow \sin \theta=\frac{\sqrt{14}}{\sqrt{95}}$

$\therefore|\overrightarrow{ b }| \times|\overrightarrow{ a } \times \overrightarrow{ c }|=3 \sqrt{95}$

$\Rightarrow|\overrightarrow{ a } \times \overrightarrow{ c }|=\sqrt{3} \times \sqrt{95}$

$|\vec{a}|=|\overrightarrow{ b } \times \overrightarrow{ c }|=|\overrightarrow{ b }||\overrightarrow{ c }| \sin \theta$

$\sqrt{11}=\sqrt{50} \sqrt{22} \sin \theta$

$\Rightarrow \sin \theta=\frac{1}{10}$

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

$=|\overrightarrow{ b }|^2+|\overrightarrow{ c }|^2+2|\overrightarrow{ b }||\overrightarrow{ c }| \cos \theta$

$=50+22+2 \times \sqrt{50} \times \sqrt{22} \times \frac{\sqrt{99}}{10}$

$=72+66$

$|72-| \vec{b}+\left.\vec{c}\right|^2 \mid=66$

$\overrightarrow{ v } \times \overrightarrow{ w }=\overrightarrow{ u }+\lambda \overrightarrow{ v } \ldots \ldots \ldots \ldots \ldots \ldots \ldots(1)$

Taking dot with $\overrightarrow{ w }$ in (1)

$\overrightarrow{ w } \cdot(\overrightarrow{ v } \times \overrightarrow{ w })=\overrightarrow{ u } \cdot \overrightarrow{ w }+\lambda \overrightarrow{ v } \cdot \overrightarrow{ w }$

$\Rightarrow 0=\overrightarrow{ u } \cdot \overrightarrow{ w }+2 \lambda$

Taking dot with $\vec{v}$ in $(1)$

$\overrightarrow{ v } \cdot(\overrightarrow{ v } \times \overrightarrow{ w })=\overrightarrow{ u } \cdot \overrightarrow{ v }+\lambda \overrightarrow{ v } \cdot \overrightarrow{ v }$

$\Rightarrow 0=(2-1+2)+\lambda(6)$

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

$\Rightarrow \overrightarrow{ u } \cdot \overrightarrow{ w }=-2 \lambda=1$

$\lambda(10)=\mu(2)+4(-14)=0$

$10 \lambda-2 \mu=56$

$5 \lambda-\mu=28$

$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\sqrt{54}$

$\frac{-2 \lambda+4 \mu-8}{\sqrt{24}}=\sqrt{54}$

$-2 \lambda+4 \mu-8=\sqrt{54 \times 24}$

By solving equation $(1)$ and $(2)$

$\Rightarrow \lambda+\mu=24$

$\overline{A C}=2 \bar{a}+3 \bar{b}-4 \bar{c}$

$\overline{A D}=\bar{a}-3 \bar{b}+5 \bar{c}$

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

$\Rightarrow(\lambda-1)(15-12)+2(-10+4)+3(6-3)=0$

$\Rightarrow(\lambda-1)=1 \Rightarrow \lambda=2$

$(2 \hat{i}+a \hat{j}+4 \hat{ k }) \cdot[( x -2) \hat{ i }+( y - a ) \hat{ j }+( z -4) \hat{ k }]=0$

$\Rightarrow 2 x + ay +4 z =20+ a ^2$

$\Rightarrow A \equiv\left(\frac{20+ a ^2}{2}, 0,0\right)$

$B \equiv\left(0, \frac{20+ a ^2}{ a }, 0\right)$

$C \equiv\left(0,0, \frac{20+ a ^2}{4}\right)$

$\Rightarrow \text { Volume of tetrahedron }$

$=\frac{1}{6}[\overrightarrow{ a } \overrightarrow{ b } \overrightarrow{ c }]$

$=\frac{1}{6} \overrightarrow{ a } .(\overrightarrow{ b } \times \overrightarrow{ c })$

$\Rightarrow\left(20+ a ^2\right)^3=144 \times 48 \times a$

$\Rightarrow a =2$

$\Rightarrow \text { Equation of plane is } 2 x +2 y +4 z =24$

$\text { Or } x + y +2 z =12$

$\Rightarrow(3,0,4) \quad \text { Not lies on the Plane }$

$x + y +2 z =12$

$\min .(|u||\vec{v} \times \vec{w}| \cos \theta)=-\alpha \sqrt{3401}$

$\cos \theta=-1$

$|u|=\alpha \text { (Given) }$

$|\vec{v} \times \vec{w}|=\sqrt{3401}$

$\vec{v} \times \vec{w}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -3 \\ 2 \alpha & 1 & -1\end{array}\right|$

$\vec{v} \times \vec{w}=\hat{i}-5 \alpha \hat{j}-3 \alpha \hat{k}$

$|\vec{v} \times \vec{w}|=\sqrt{1+25 \alpha^2+9 \alpha^2}=\sqrt{3401}$

$34 \alpha^2=3400$

$\alpha^2=100$

$\alpha=10 \quad(\text { as } \alpha > 0)$

So $\vec{u} =\lambda(\hat{i}-5 \alpha \hat{j}-3 \alpha \hat{k})$

$\vec{u} =\sqrt{\lambda^2+25 \alpha^2 \lambda^2+9 \alpha^2 \lambda}$

$\alpha^2 =\lambda^2\left(1+25 \alpha^2+9 \alpha^2\right)$

$100 =\lambda^2(1+34 \times 100)$

$\lambda^2 =\frac{100}{3401}=\frac{m}{n}$

Hence $\left[\begin{array}{lll}\overrightarrow{ BA } & \overrightarrow{ CA } & \overrightarrow{ DA }\end{array}\right]=0$

$\begin{aligned}& \left|\begin{array}{ccc}4 & 3 & 2 \lambda-3 \\7 & 5-\lambda & 2 \lambda-4 \\6 & 0 & 2 \lambda-6\end{array}\right|=0 \\& \lambda=2,3 \text { Hence } \sum_{\lambda \in S}(\lambda+2)^2=41\end{aligned}$

$\Rightarrow b -1+ c -1+ a (1- bc )=0$

$\therefore abc = a + b + c -2$

$\left[\overrightarrow{ v }_1 \overrightarrow{ v }_2 \overrightarrow{ v }_3\right]=0 \quad \therefore\left|\begin{array}{ccc} a + b & c & c \\ a & b + c & a \\ b & b & c + a \end{array}\right|=0$

$R_3 \rightarrow R_3-\left(R_1+R_2\right) \Rightarrow\left|\begin{array}{ccc}a+b & c & c \\ a & b+c & a \\ -2 a & -2 c & 0\end{array}\right|=0$

$\Rightarrow-2 a\left(a c-b c-c^2\right)+2 c\left(a^2+a b-a c\right)=0$

$\Rightarrow-2 a^2 c+2 a b c+2 a c^2+2 a^2 c+2 a b c-2 a c^2=0$

$\Rightarrow 4 a b c=0 \quad \therefore a b c=0$

$\therefore a + b + c =2 \quad \therefore 6( a + b + c )=12$ Ans.

$\overrightarrow{ b }=3 \hat{ i }+5 \hat{ k }$

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

$\overrightarrow{ d }=\lambda(\overrightarrow{ a } \times \overrightarrow{ b })=\lambda\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 7 & -1 \\ 3 & 0 & 5\end{array}\right|$

$\overrightarrow{ d }=\lambda(35 \hat{ i }-13 \hat{ j }-21 \hat{ k })$

$\lambda(35+13-42)=12$

$\lambda=2$

$\overrightarrow{ d }=2(35 \hat{ i }-13 \hat{ j }-21 \hat{ k })$

$(\hat{ i }+\hat{ j }-\hat{ k })(\overrightarrow{ c } \times \overrightarrow{ d })$

$=\left|\begin{array}{ccc}-1 & 1 & -1 \\ 1 & -1 & 2 \\ 70 & -26 & -42\end{array}\right|=44$

$\lambda(10+4 \mu)+(5-3 \mu)+(-10)=0$

$(\mu+5)(4 \mu+10)+5-3 \mu-10=0$

$\mu=-15 ; \lambda=5 / 4$

$\mu=-3 ; \lambda=2$

$\text { Hence } \sum_{(\lambda, \mu) \in S} 80\left(\lambda^2+\mu^2\right)$

$=80\left(\frac{250}{16}+13\right)$

$=1250+1040$

$=2290$

$\overrightarrow{ b } \cdot \overrightarrow{ c }=12$

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

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

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

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

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

$=12 \times(|\overrightarrow{ n }|)$

$=12 \times 1$

$=12$

Now $\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1$

$=(\hat{ i }+2 \hat{ j }-4 \hat{ k })-\lambda(4 \hat{ i }+3 \hat{ j }+5 \hat{ k })$

$=(1-4 \lambda) \hat{ i }+(2-3 \lambda) \hat{ j }-(5 \lambda+4) \hat{ k }$

$\vec{\beta}_2 \cdot \vec{\alpha}=0$

$\Rightarrow 4(1-4 \lambda)+3(2-3 \lambda)-5(5 \lambda+4)=0$

$\Rightarrow 4-16 \alpha+6-9 \lambda-25 \lambda-20=0$

$\Rightarrow 50 \lambda=-10$

$\Rightarrow \lambda=\frac{-1}{5}$

$\vec{\beta}_2=\left(1+\frac{4}{5}\right) \hat{ i }+\left(2+\frac{3}{5}\right) \hat{ j }-(-1+4) \hat{ k }$

$\vec{\beta}_2=\frac{9}{5} \hat{ i }+\frac{13}{5} \hat{ j }-3 \hat{ k }$

$5 \vec{\beta}_2=9 \hat{ i }+13 \hat{ j }-15 \hat{ k }$

$5 \vec{\beta}_2 \cdot(\hat{ i }+\hat{ j }+\hat{ k })=9+13-15=7$

$(\vec{u}) \cdot(\alpha \vec{u}+\beta \vec{v})=0$

$\vec{u} \cdot \vec{v}=|u||v| \cos 45^{\circ}$

$\alpha=-\frac{\beta}{\sqrt{2}}$

$=|\alpha \vec{u}+\beta \vec{v}|=r$

$\alpha^2+\beta^2+\sqrt{2} \alpha \beta=1$

$\alpha=-1, \beta^2=2$

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

$\overrightarrow{ c }=\alpha(\overrightarrow{ a }+\overrightarrow{ b })=\alpha(\lambda+3) \hat{ i }+\alpha \hat{ k }$

$\vec{b} \cdot \vec{c}=-20 \Rightarrow 3 \alpha(\lambda+3)+2 \alpha=-20$

$\vec{a} . \vec{c}=-17 \Rightarrow \alpha \lambda(\lambda+3)-\alpha=-17$

$\Rightarrow \alpha(3 \lambda+9+2)=-20$

$\alpha\left(\lambda^2+3 \lambda-1\right)=-17$

$17(3 \lambda+11)=20\left(\lambda^2+3 \lambda-1\right)$

$20 \lambda^2+9 \lambda-207=0$

$\lambda=3 \quad(\lambda \in Z)$

$\Rightarrow \alpha=-1 \quad \Rightarrow \overrightarrow{ c }=-(6 \hat{ i }+\hat{ k })$

$\overrightarrow{ v }=\overrightarrow{ c } \times(3 \hat{ i }+\hat{ j }+\hat{ k })$

$=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ -6 & 0 & -1 \\ 3 & 1 & 1\end{array}\right|=\hat{i}+3 \hat{j}-6 \hat{k}$

$|\overrightarrow{ v }|^2=(-1)^2+3^2+6^2=46$

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

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

$\text { Also } \overrightarrow{ d } \cdot \overrightarrow{ a }=24$

$\Rightarrow(\overrightarrow{ c }+\lambda \overrightarrow{ b }) \cdot \overrightarrow{ a }=24$

$\lambda=\frac{24-\overrightarrow{ a } \cdot \overrightarrow{ c }}{\overrightarrow{ b } \cdot \overrightarrow{ a }}=\frac{24-6}{9}=2$

$\Rightarrow \overrightarrow{ d }=\overrightarrow{ c }+2(\overrightarrow{ b })$

$\qquad=8 \hat{ i }-5 \hat{ j }+18 \hat{ k }$

$\Rightarrow|\overrightarrow{ d }|^2=64+25+324=413$

$\vec{v}=\lambda(1,1,2)+\mu(2,-3,1)$

$\vec{v}=(\lambda+2 \mu, \lambda-3 \mu, 2 \lambda+\mu)$

$\overrightarrow{ v } \cdot \hat{ j }=7$

$\lambda-3 \mu=7$

$\overrightarrow{ v } \cdot \frac{\overrightarrow{ c }}{|\overrightarrow{ c }|}=\frac{2}{\sqrt{3}}$

$\overrightarrow{ V } \cdot \overrightarrow{ C }=2$

$\lambda+2 \mu-\lambda+3 \mu+2 \lambda+\mu=2$

$2 \lambda+6 \mu=2$

$\lambda+3 \mu=1$

$\lambda-3 \mu=7$

$2 \lambda=8$

$\lambda=4$

$\mu=-1$

We get $\quad \overrightarrow{ v }=(2,7,7)$

$\Rightarrow \frac{\alpha+6+2}{\sqrt{1+4+4}}=\frac{10}{3} \Rightarrow \alpha=2$

and

$\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 3 & -\beta & 4 \\ 1 & 2 & -2\end{array}\right|=-6 \hat{ i }+\hat{ j }+\hat{ k }$

$\Rightarrow 2 \beta-8=-6 \Rightarrow \beta=1$

$\Rightarrow \alpha+\beta=3$

$\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$

$\vec{a} \cdot \vec{b}=3$

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

$5+9=6|\vec{b}|^{2}$

$|b|^{2}=\frac{7}{3}$

$|\vec{a}-\vec{b}|=\sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b}}=\sqrt{\frac{7}{3}}$

projection of $\vec{b}$ on $\vec{a}-\vec{b}=\frac{\vec{b} \cdot(\vec{a}-\vec{b})}{|\vec{a}-\vec{b}|}$

$=\frac{\vec{b} \cdot \vec{a}-|\vec{b}|^{2}}{|\vec{a}-\vec{b}|}=\frac{3-\frac{7}{3}}{\sqrt{\frac{7}{3}}}$

$=\frac{2}{\sqrt{21}}$

Projection of $\vec{a} \times \vec{b}$ on $-\hat{i}+2 \hat{j}-2 \hat{k}$

$=\frac{(\vec{a} \times \vec{b}) \cdot(-\hat{i}+2 \hat{j}-2 \hat{k})}{3}=30$

$2 \alpha^{2}-\alpha-91=0$

$\alpha=7,-\frac{13}{2}$

$|(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})|^{2}+4(\vec{a} \cdot \vec{b})^{2}$

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

$2|\vec{a} \times \vec{b}|^{2}+4 a^{2} b^{2} \cos ^{2} \theta$

$4 a^{2} b^{2} \sin ^{2} \theta+4 a^{2} b^{2} \cos ^{2} \theta$

$4 a^{2} b^{2}=4 \times 16 \times 9$

$=576$

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

$\vec{b} \times \vec{c}=\vec{a}$

$|\vec{b}| \in\{1,2 \ldots \ldots 10\}$

$\because \vec{b} \times \vec{c}=\vec{a}$

$\Rightarrow \vec{a}$ is perpendicular to $\vec{b}$ as well as $\vec{a}$ is perpendicular to $\vec{C}$

Now $\vec{a} \cdot \vec{c}=2-3-2=-3 \neq 0$

This $\vec{b} \times \vec{c}=\vec{a}$ is not possible.

No. of vectors $\vec{b}=0$

$\overline{ b } \cdot \overline{ c }=0 \Rightarrow 3 c _{1}+3 c _{2}+ c _{3}=0$.............$(2)$

And $[\overline{ a } \overline{ b } \overline{ c }]=0 \Rightarrow\left|\begin{array}{ccc} c _{1} & c _{2} & c _{3} \\ 2 & 1 & 3 \\ 3 & 3 & 1\end{array}\right|=0$

$\Rightarrow 8 c_{1}-7 c_{2}-3 c_{3}=0$..............$(3)$

By solving $(1), (2), (3)$ we get

$c _{1}=\frac{10}{122}, c _{2}=\frac{-85}{122}, c _{3}=\frac{225}{122}$

$\therefore 122\left( c _{1}+ c _{2}+ c _{3}\right)=150$

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

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

$\Rightarrow \vec{a} \cdot \vec{c}=1, \vec{a} \cdot \vec{b}=-\lambda$

$\Rightarrow(3 \hat{ i }+\hat{ j }) \cdot(\hat{ i }+2 \hat{ j }+\hat{ k })=-\lambda$

$\Rightarrow \lambda=-5$

$\Rightarrow 13-1-4 \lambda=0 \Rightarrow \lambda=3$

$\Rightarrow \overrightarrow{ b }=\hat{ i }+\hat{ j }+3 \hat{ k } \Rightarrow \overrightarrow{ a } \times \overrightarrow{ b }=13 \hat{ i }-\hat{ j }-4 \hat{ k }$

$\Rightarrow(\overrightarrow{ a } \times \overrightarrow{ b }) \times \overrightarrow{ b }=(13 \hat{ i }-\hat{ j }-4 \hat{ k }) \times(\hat{ i }+\hat{ j }+3 \hat{ k })$

$\Rightarrow-21 \overrightarrow{ b }-11 \overrightarrow{ a }=\hat{ i }-43 \hat{ j }+14 \hat{ k }$

$\Rightarrow \overrightarrow{ a }=-2 \hat{ i }+2 \hat{ j }-7 \hat{ k }$

Now $(\overrightarrow{ b }-\overrightarrow{ a }) \cdot(\hat{ k }-\hat{ j })+(\overrightarrow{ b }+\overrightarrow{ a }) \cdot(\hat{ i }-\hat{ k })=14$

$=(2 \hat{i}-13 \hat{j}-4 \hat{k}) \times\left(3 \hat{i}+\frac{1}{2} \hat{j}+2 \hat{k}\right)$

$-(6+2) \overrightarrow{ a }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & -13 & -4 \\ 3 & \frac{1}{2} & 2\end{array}\right|$

$\vec{a}=3 \hat{i}+2 \hat{j}-5 \hat{k}$

Projection of $\vec{a}$ on vector $2 \hat{i}+2 \hat{j}+\hat{k}$ is

$\overrightarrow{ a } \cdot \frac{(2 \hat{ i }+2 \hat{ j }+\hat{ k })}{3}=\frac{5}{3}$

$\vec{a}, \vec{b}, \vec{c}$ are non-collinear for $\alpha=2$ (smallest positive integer)

Mid-point of $BC = M \left(\frac{5}{2}, 0, \frac{9}{2}\right)$

$AM =\sqrt{\frac{9}{4}+16+\frac{9}{4}}=\frac{\sqrt{82}}{2}$

$\overrightarrow{ a } \times \overrightarrow{ b }+|\overrightarrow{ b }|^{2} \overrightarrow{ c }-5 \overrightarrow{ b }=0$

It gives $\overrightarrow{ c }=\frac{1}{3}(10 \hat{ i }+3 \hat{ j }+2 \hat{ k })$

so $3 \vec{a} \cdot \vec{c}=10$

But it does not satisfy $\vec{a}+\vec{b} \times \vec{c}=0$.

This question has data error.

Alternate (Explanation) :

According to given $\vec{a}$ and $\vec{b}$

$\overrightarrow{ a } \cdot \overrightarrow{ b }=1-2+3=2 \ldots (i)$

but given equation

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

$\Rightarrow \overrightarrow{ a } \perp \overrightarrow{ b } \Rightarrow \overrightarrow{ a } \cdot \overrightarrow{ b }=0$

which contradicts.

$\overrightarrow{ a }=\lambda\left(\frac{1}{\sqrt{3}} \hat{ i }+\frac{1}{\sqrt{3}} \hat{ j }+\frac{1}{\sqrt{3}} \hat{ k }\right)=\frac{\lambda}{\sqrt{3}}(\hat{ i }+\hat{ j }+\hat{ k }$

Now projection of $\vec{a}$ on $\vec{b}=7$

$\Rightarrow \frac{\overrightarrow{ a } \cdot \overrightarrow{ b }}{|\overrightarrow{ b }|}=7$

$\frac{\lambda}{\sqrt{3}} \frac{(\hat{ i }+\hat{ j }+\hat{ k }) \cdot(3 \hat{ i }+4 \hat{ j })}{5}=7$

$\lambda=5 \sqrt{3}$

$\overrightarrow{ a }=5(\hat{ i }+\hat{ j }+\hat{ k })$

now $\overrightarrow{ b }=5 \alpha(\hat{ i }+\hat{ j }+\hat{ k })+\beta(\hat{ i })$

$\overrightarrow{ a } \cdot \overrightarrow{ b }=0$

$\Rightarrow 25 \alpha(3)+5 \beta=0$

$\Rightarrow 15 \alpha+\beta=0 \Rightarrow \beta=-15 \alpha$

$\overrightarrow{ b }=5 \alpha(-2 \hat{ i }+\hat{ j }+\hat{ k })$

$|\vec{b}|=5 \sqrt{3}$

$\Rightarrow \alpha=\pm \frac{1}{\sqrt{2}}$

$\vec{b}=\pm \frac{5}{\sqrt{2}}(-2 \hat{i}+\hat{j}+\hat{k})$

$\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}$

$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & \beta \\ 3 & -5 & 4\end{array}\right|$

$\Rightarrow(4+5 \beta) \hat{i}+(3 \beta-4 \alpha) \hat{j}+(-5 \alpha-3) \hat{k}$

$=-\hat{i}+9 \hat{j}+12 \hat{k}$

$\therefore 4+5 \beta=-1,3 \beta-4 \alpha=9,-5 \alpha-3=12$

$\beta=-1, \quad \alpha=-3$

$\therefore \vec{a}=-3 \hat{i}+\hat{j}-\hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$

$\therefore \vec{a}+\vec{b}=-4 \hat{j}+3 \hat{k}$

$|\vec{a}|^{2}=11,|\vec{b}|^{2}=50$

$\vec{a} \cdot \vec{b}=-9+(-5)-4=-18$

$\therefore$ Projectile of $(\vec{b}-2 \vec{a})$ on $\vec{a}+\vec{b}$ is

$\frac{(\vec{b}-2 \vec{a}) \cdot(\vec{a}+\vec{b})}{|\vec{a}+\vec{b}|}$

$=\frac{|\vec{b}|^{2}-2|\vec{a}|^{2}-(\vec{a} \cdot \vec{b})}{|\vec{a}+\vec{b}|}=\frac{50-22-(-18)}{5}=\frac{46}{5}$

Ans. $\left(\frac{46}{5}\right)$

$\overrightarrow{ u } \cdot \overrightarrow{ v }>0$

$a \left(\log _{ e } b \right)^{2}-12+6 a \left(\log _{ e } b \right)>0$

$\forall b >1$

let $\log _{ e } b = t \Rightarrow t >0$ as $b >1$

$y =a at ^{2}+6 a at -12 \& y >0, \forall t >0$4

$a \in \phi$

$\Rightarrow 6 x y|\bar{a}|^{2}-18 x^{2}(\bar{a} \cdot \bar{b})+6 y^{2}(\bar{a} \cdot \bar{b})-18 x y|\bar{b}|^{2}=0$

$\Rightarrow 6 x y\left(|\vec{a}|^{2}-3|\bar{b}|^{2}\right)+(\bar{a} \cdot \bar{b})\left(y^{2}-3 x^{2}\right)=0$

This should hold $\forall x , y \in R \times R$

$\therefore|\overline{ a }|^{2}=3|\overline{ b }|^{2} \&(\overline{ a } \cdot \overline{ b })=0$

Now $|\bar{a} \times \bar{b}|^{2}=|\bar{a}|^{2}|\bar{b}|^{2}-(\bar{a} \cdot \bar{b})^{2}$

$=|\overline{ a }|^{2} \cdot \frac{|\overline{ a }|^{2}}{3}$

$\therefore \&|\overline{ a } \times \overline{ b }|$ $=\frac{|\overline{ a }|^{2}}{\sqrt{3}}$ $=\frac{81}{\sqrt{3}}=27 \sqrt{3}$

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

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

$|\vec{a} \times \vec{b}|^{2}=75$

$|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2}=75$

$6|\vec{a}|^{2}-9=75 \Rightarrow|\vec{a}|^{2}=14$

$\hat{a} \cdot \hat{b}=|\hat{a}||\hat{b}| \cos \phi$

$\hat{a} \cdot \hat{b}=\cos \phi=\frac{1}{\sqrt{2}}$

$\cos \theta=\frac{(\hat{a}+\hat{b}) \cdot(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))}{|\hat{a}+\hat{b}||\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b})|}$

$|\hat{a}+\hat{b}|^{2}=(\hat{a}+\hat{b}) \cdot(\hat{a}+\hat{b})$

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

$=2+\sqrt{2}$

$\hat{a} \times \hat{b}=|\hat{a}||\hat{b}| \sin \phi \hat{n}$

$\hat{ a } \times \hat{ b }=\frac{\hat{ n }}{\sqrt{2}} \quad$ when $\hat{ n }$ is vector $\perp \hat a$ and $\hat{ b }$

let $\vec{c}=\hat{a} \times \hat{b}$

We know.

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

$\overrightarrow{ c } \cdot \overrightarrow{ b }=0$

$|\hat{a}+2 \hat{b}+2 \vec{c}|^{2}$

$=1+4+\frac{(4)}{2}+4 \hat{ a } \cdot \hat{ b }+8 \hat{ b } \cdot \overrightarrow{ c }+4 \overrightarrow{ c } \cdot \hat{ a }$

$=7+\frac{4}{\sqrt{2}}=7+2 \sqrt{2}$

Now,$(\hat{a}+\hat{b}) \cdot(\hat{a}+2 \hat{b}+2 \vec{c})$

$=|\hat{a}|^{2}+2 \hat{a} \cdot \hat{b}+0+\hat{b} \cdot \hat{a}+2|\hat{b}|^{2}+0$

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

$=3+\frac{3}{\sqrt{2}}$

$\cos \theta=\frac{3+\frac{3}{\sqrt{2}}}{\sqrt{2+\sqrt{2}} \sqrt{7+2 \sqrt{2}}}$

$\cos ^{2} \theta=\frac{9(\sqrt{2}+1)^{2}}{2(2+\sqrt{2})(7+2 \sqrt{2})}$

$\cos ^{2} \theta=\left(\frac{9}{2 \sqrt{2}}\right) \frac{(\sqrt{2}+1)}{(7+2 \sqrt{2})}$

$164 \cos ^{2} \theta=\frac{(82)(9)}{\sqrt{2}} \frac{(\sqrt{2}+1)}{(7+2 \sqrt{2})} \frac{(7-2 \sqrt{2})}{(7-2 \sqrt{2})}$

$=\frac{(82)}{\sqrt{2}} \frac{(9)[7 \sqrt{2}-4+7-2 \sqrt{2}]}{(41)}$

$=(9 \sqrt{2})[5 \sqrt{2}+3]$

$=90+27 \sqrt{2}$

$|\hat{ b }|^{2}=| c |^{2}+4|\overrightarrow{ c } \times \hat{ a }|^{2}+4 \overrightarrow{ c } \cdot(\overrightarrow{ c } \times \hat{ a })$

$1=|c|^{2}+4|c|^{2} \sin ^{2} \frac{\pi}{12}+0$

$1=|c|^{2}+4|c|^{2}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)^{2}$

$|c|^{2}=\frac{1}{3-\sqrt{3}}=\frac{3+\sqrt{3}}{6}$

So $6^{2}|c|^{2}=6(3+\sqrt{3})$

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

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

${[3 \overrightarrow{ b }-\overrightarrow{ c }, \overrightarrow{ c }-2 \overrightarrow{ a }, 3 \overrightarrow{ b }-2 \overrightarrow{ a }]}$

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

$(3 \overrightarrow{ b }-\overrightarrow{ c }) \cdot[3(\overrightarrow{ c } \times \overrightarrow{ b })-2(\overrightarrow{ c } \times \overrightarrow{ a })-6(\overrightarrow{ a } \times \overrightarrow{ b })]$

$-6[\overrightarrow{ b } \overrightarrow{ c } \overrightarrow{ a }+6[\overrightarrow{ c } \overrightarrow{ a } \overrightarrow{ b }]$

$[\overline{ a } \overline{ b } \overline{ c }] \neq 0$

Now, $[\bar{a} \bar{b} \bar{c}]$

$=\left|\begin{array}{ccc}1+ t & 1- t & 1 \\ 1- t & 1+ t & 2 \\ t & - t & 1\end{array}\right|$

$C _{2} \rightarrow C _{1}+ C _{2}$

$\left|\begin{array}{ccc}1+ t & 2 & 1 \\ 1- t & 2 & 2 \\ t & 0 & 1\end{array}\right|$

$=2\left|\begin{array}{ccc}1+ t & 1 & 1 \\ 1- t & 1 & 2 \\ t & 0 & 1\end{array}\right|$

$=2[(1+ t )-(1- t )+ t ]$

$=2[3 t ]=6 t$

$[\overline{ a } \overline{ b } \overline{ c }] \neq 0 \Rightarrow t \neq 0$

$[\overline{ AB } \overline{ AC } \overline{ AD }]=0$

$\left|\begin{array}{ccc}3 & -1 & -8 \\ -1 & \lambda-3 & -1 \\ \lambda-2 & -1 & -6\end{array}\right|=0$

$\Rightarrow[-6(\lambda-3)-1]-8(1-(\lambda-3)(\lambda-2))+(6+(\lambda$

$-2)=0$

$3(-6 \lambda+17)-8\left(-\lambda^{2}+5 \lambda-5\right)+(\lambda+4)=8$

$8 \lambda^{2}-57 \lambda+95=0$

$\lambda_{1} \lambda_{2}=\frac{95}{8}$

$((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k}=\frac{23}{2}$, then $|\vec{b} \times 2 \hat{j}|$ is

$((\vec{a} \hat{i}) \vec{b}-(\vec{b} \hat{i}) \vec{a}) \cdot \hat{k}=\frac{23}{2}$

$(\vec{a} \cdot \hat{i})(\vec{b} \cdot \hat{i})-(\vec{b} \hat{i})(\vec{a} \cdot \hat{k})=\frac{23}{2}$

$2 \times 2-\alpha \times 5=\frac{23}{2} \Rightarrow 5 \alpha=4-\frac{23}{2} \Rightarrow \alpha=\frac{-3}{2}$

$\vec{b} \times 2 \hat{j}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \alpha & \beta & 2 \\ 0 & 2 & 0 \end{array}\right|=-4 \hat{i}+2 \alpha \hat{k}$

$\therefore|\vec{b} \times 2 \hat{j}|=\sqrt{16+4 \alpha^{2}}=\sqrt{16+4 \times \frac{9}{4}}=5$

$|\overrightarrow{ a }|=1$ and $|\overrightarrow{ a } \cdot \overrightarrow{ b }|=|\overrightarrow{ a } \times \overrightarrow{ b }|$

$\Rightarrow \cos \theta=\sin \theta$

$\Rightarrow \theta=\frac{\pi}{4}$

$\therefore|\overrightarrow{ a } \times \overrightarrow{ b }|=4 \sqrt{2}\Rightarrow|\vec{a}||\overrightarrow{ b }| \sin \frac{\pi}{4}=4 \sqrt{2}$

$\Rightarrow|\overrightarrow{ b }|=8$

Now, $\overrightarrow{ c }=2 \sqrt{2}(\overrightarrow{ a } \times \overrightarrow{ b })-2 \overrightarrow{ b }$

$|\overrightarrow{ c }|=\sqrt{(2 \sqrt{2})^{2}|\overrightarrow{ a } \times \overrightarrow{ b }|^{2}+(2|\overrightarrow{ b }|)^{2}}=16 \sqrt{2}$

Now, $\overrightarrow{ b } \cdot \overrightarrow{ c }=-2|\overrightarrow{ b }|^{2}$

$\Rightarrow 8 \times 16 \sqrt{2} \times \cos \alpha=-2.64$

$\Rightarrow \cos \alpha=-\frac{1}{\sqrt{2}} \Rightarrow \alpha=\frac{3 \pi}{4}$

$\overrightarrow{ b }+\overrightarrow{ c }=-\overrightarrow{ a }$

$|\overrightarrow{ b }|^{2}+|\overrightarrow{ c }|^{2}+2 \overrightarrow{ b } \cdot \overrightarrow{ c }=|\overrightarrow{ a }|^{2}$

$|\overrightarrow{ c }|^{2}=36$

$|\overrightarrow{ c }|=6$

$S 1:|\overrightarrow{ a } \times \overrightarrow{ b }+\overrightarrow{ c } \times \overrightarrow{ b }|-|\overrightarrow{ c }|$

$|(\overrightarrow{ a }+\overrightarrow{ c }) \times \overrightarrow{ b }|-|\overrightarrow{ c }|$

$|-\overrightarrow{ b } \times \overrightarrow{ b }|-|\overrightarrow{ c }|$

$0-6=-6$

$S 2: \overrightarrow{ a }+\overrightarrow{ b }+\overrightarrow{ c }=0$

$\overrightarrow{ b }+\overrightarrow{ c }=-\overrightarrow{ a }$

$|\overrightarrow{ a }|^{2}+|\overrightarrow{ b }|^{2}-2|\overrightarrow{ a }||\overrightarrow{ b }| \cos (\angle ACB )=|\overrightarrow{ c }|^{2}$

$\cos (\angle ACB )=\sqrt{\frac{2}{3}}$

$\overrightarrow{ n }_{2}=(\hat{ i }+\hat{ k }) \times(\hat{ i }-\hat{ j })$

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

Line of intersection along $\overrightarrow{ n }_{1} \times \overrightarrow{ n }_{2}$

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

D.R of $\vec{a}=-\hat{i}+\hat{j}$

D.R of $\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}$

$\vec{a} \cdot \vec{b}=-3$ and $(\vec{a} \wedge \vec{b})=\theta$

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

$\theta=\frac{3 \pi}{4}$

$((\hat{a}+\hat{b})+2(\hat{a} \times \hat{b})) \cdot((\hat{a}+\hat{b})+2(\hat{a} \times \hat{b}))=4$

$|\hat{a}+\hat{b}|^{2}+4|(\hat{a} \times \hat{b})|^{2}+0=4$

Let the angle be $\theta$ between $\hat{a}$ and $\hat{b}$

$2+2 \cos \theta+4 \sin ^{2} \theta=4$

$2+2 \cos \theta-4 \cos ^{2} \theta=0$

Let $\cos \theta= t$ then

$2 t ^{2}- t -1=0$

$2 t ^{2}-2 t + t -1=0$

$2 t ( t -1)+( t -1)=0$

$(2 t +1)( t -1)=0$

$t =-\frac{1}{2} \quad \text { or } \quad t =1$

$\cos \theta=-\frac{1}{2}$           

$\theta=\frac{2 \pi}{3}$

not possible as $\theta \in(0, \pi)$

$S_{1} \quad 2|\vec{a} \times \vec{b}|=2 \sin \left(\frac{2 \pi}{3}\right)$

$|\hat{a}-\hat{b}| =\sqrt{1+1-2 \cos \left(\frac{2 \pi}{3}\right)}$

$=\sqrt{2-2 \times\left(-\frac{1}{2}\right)}$

$=\sqrt{3}$

$S_{1}$ is correct.

$S_{2}$ projection of $\hat{a}$ on $(\hat{a}+\hat{b})$.

$\frac{\hat{a} .(\hat{a}+\hat{b})}{|\hat{a}+\hat{b}|}=\frac{1+\cos \left(\frac{2 \pi}{3}\right)}{\sqrt{2+2 \cos \frac{2 \pi}{3}}}$

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

$=\frac{1}{2}$

area of parallelogram $=|\hat{a} \times \hat{b}|$

$|\hat{a} \times \hat{b}|=\sqrt{(\alpha+2)^{2}+(\alpha-2)^{2}+\left(\alpha^{2}+4\right)^{2}}$

Given $|\hat{a} \times \hat{b}|=\sqrt{15\left(\alpha^{2}+4\right)}$

$2\left(\alpha^{2}+4\right)+\left(\alpha^{2}+4\right)^{2}=15\left(\alpha^{2}+4\right)$

$\left(\alpha^{2}+4\right)^{2}=13\left(\alpha^{2}+4\right)$

$\Rightarrow \alpha^{2}+4=13 \therefore \alpha^{2}=9$

$2|\vec{a}|^{2}+(\vec{a} \cdot \vec{b})|\vec{b}|^{2}$

$|\vec{a}|^{2}=\alpha^{2}+4+1=\alpha^{2}+5$

$|\vec{b}|^{2}=4+\alpha^{2}+1=\alpha^{2}+5$

$\vec{a} \cdot \vec{b}=-2 \alpha+2 \alpha-1=-1$

$\therefore 2|\vec{a}|^{2}+(\vec{a} \cdot \vec{b})|\vec{b}|^{2}$

$2\left(\alpha^{2}+5\right)-1\left(\alpha^{2}+5\right)=\alpha^{2}+5=14$

$\overrightarrow{ a }^{\wedge} \overrightarrow{ b }=\overrightarrow{ b } \wedge \overrightarrow{ c }=\overrightarrow{ c }^{\wedge} \overrightarrow{ a }=\theta=\frac{2 \pi}{3}$

So, $\vec{a} \cdot \vec{b}=-\frac{1}{2} a b, \vec{b} \cdot \vec{c}=-\frac{1}{2} b c, \vec{a} \cdot \vec{c} .=-\frac{1}{2} a c$ (let)

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

$=\frac{1}{4} a b^{2} c+\frac{1}{2} a b^{2} c=\frac{3}{4} a b^{2} c$

Similarly

$(\overrightarrow{ b } \times \overrightarrow{ c }) \cdot(\overrightarrow{ c } \times \overrightarrow{ a })=\frac{3}{4} abc ^{2}$

$(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=\frac{3}{4} a^{2} b c$

$168=\frac{3}{4} a b c(a+b+c)$

So, $(a+b+c)=16$