Mathematics M.C.Q (1 Marks)

$\frac{x-2}{(-3)}=\frac{y-\frac{2}{3}}{\left(\frac{4 \lambda+1}{3}\right)}=\frac{z-4}{(-1)}$

$ \frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7} $...................($2$)

$ \frac{x+3}{3 \mu}=\frac{y-\frac{1}{2}}{(-3)}=\frac{z-5}{(-7)} $

 Right angle $ \Rightarrow(-3)(3 \mu)+\left(\frac{4 \lambda+1}{3}\right)(-3)+(-1)(-7)=0 $

$ -9 \mu-4 \lambda-1+7=0 $

$ 4 \lambda+9 \mu=6$

$ \mathrm{P}(-3 \lambda-2,4 \lambda+2,2 \lambda+5) $

$ \mathrm{Q}(-\mu-2,2 \mu-6,1) $

$ \mathrm{DRS} \text { of } \mathrm{PQ}=(3 \lambda-\mu, 2 \mu-4 \lambda-8,-2 \lambda-4)$

$D R S$ of $P Q=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 0 \\ -3 & 4 & 2\end{array}\right|$

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

OR

$ (2,1,1) $

$ \frac{3 \lambda-\mu}{2}=\frac{2 \mu-4 \lambda-8}{1}=\frac{-2 \lambda-4}{1} $

$ \Rightarrow \mu=\lambda+2 \& 7 \lambda=\mu-8 $

$ \lambda=-1 \quad \mu=1 $

$ Q:(-3,-4,1) $

$ L_{P Q}=\frac{x+3}{2}=\frac{y+4}{1}=\frac{z-1}{1} $

$ (-1, \alpha, \beta) \Rightarrow 1=\frac{\alpha+4}{1}=\frac{\beta-1}{1} $

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

$ (\alpha-\beta)^2=25$

$\mathrm{S} . \mathrm{D}=\frac{\left|\left(\overline{\mathrm{a}}_2 \cdot \overline{\mathrm{a}}_1\right) \cdot\left(\overline{\mathrm{b}}_1 \cdot \overline{\mathrm{b}}_2\right)\right|}{\left|\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2\right|}$

$\begin{array}{ll}a_1=3,-15,9 & b_1=2,-7,5 \\ a_2=-1,1,9 & b_2=2,1,-3 \\ a_2-a_1=-4,16,0 & \end{array}$

$\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -7 & 5 \\ 2 & 1 & -3\end{array}\right|=\hat{\mathrm{i}}(16)-\hat{\mathrm{j}}(-16)+\hat{\mathrm{k}}(16)$

$ 16(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) $

$ \left|\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2\right|=16 \sqrt{3} $

$ \therefore\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right) \cdot\left(\overline{\mathrm{b}}_1-\overline{\mathrm{b}}_2\right)=16[-4+16]=(16)(12) $

$ \text { S.D. }=\frac{(16)(12)}{16 \sqrt{3}}=4 \sqrt{3}$

$ \frac{x+6}{2}=\frac{y-7}{-3}=\frac{z+5}{4}=\lambda $

$ Q(2 \lambda-6,7-3 \lambda, 4 \lambda-5) $

$ \overline{Q R}(2 \lambda-7,-3 \lambda, 4 \lambda-11) $

$ \overline{Q R} \cdot \text { dr's of line }=0 $

$ 4 \lambda-14+9 \lambda+16 \lambda-44=0 $

$ 29 \lambda=58 \Rightarrow \lambda=2 $

$ Q(-2,1,3) $

$ P Q=\sqrt{144+9+16}=\sqrt{169}=13$

$ \overrightarrow{\mathrm{QA}}=(\mathrm{t}-1) \hat{\mathrm{i}}+(2 \mathrm{t}-5) \hat{\mathrm{j}}+(3 \mathrm{t}-2) \hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{QA}} \cdot \overrightarrow{\mathrm{b}}=0 $

$ (\mathrm{t}-1)+2(2 \mathrm{t}-5)+3(3 \mathrm{t}-2)=0 $

$ 14 \mathrm{t}=17 $

$ \alpha=\frac{20}{14} \quad \beta=\frac{12}{14} \quad \gamma=\frac{102}{14} $

$ 2 \alpha+\beta+\gamma=\frac{154}{14}=11$

$\vec{n}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1\end{array}\right|=19 \hat{i}+11 \hat{j}+9 \hat{k}$

S.d. $=$ projection of $\overrightarrow{\mathrm{AB}}$ on $\vec{n}$

$=\left|\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}|}\right|=\left|\frac{(2 \hat{\mathrm{i}}+16 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \cdot(19 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+9 \hat{\mathrm{k}})}{\sqrt{361+121+81}}\right|$

$ =\frac{38+176-27}{\sqrt{563}} $

$ \text { S.d. }=\frac{187}{\sqrt{563}}$

$ \overrightarrow{\mathrm{AM}} \cdot \overrightarrow{\mathrm{b}}=0 $

$ \Rightarrow \quad 9 \lambda-15+4 \lambda-2+16 \lambda-12=0 $

$ \Rightarrow \quad 29 \lambda=29 $

$ \Rightarrow \quad \lambda=1 $

$ \mathrm{M}(4,2,6), \mathrm{I}=(2,3,7) $

$ \text { Required Distance }=\sqrt{4+9+49}=\sqrt{62}$

Ans. $62$

Direction ratio of $\mathrm{PQ}=(1,4,-1)$

Now, $\cos \theta=\left|\frac{4+4+1}{\sqrt{18} \cdot \sqrt{18}}\right|$

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

Point $B$ lies on $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$

$\frac{2 \lambda+7-6}{1}=\frac{-3 \lambda-2-4}{0}=\frac{6 \lambda+11-8}{3}$

$-3 \lambda-6=0$

$\lambda=-2$

$\text { B } \Rightarrow(3,4,-1)$

$\mathrm{AB}  =\sqrt{(7-3)^2+(4+2)^2+(11+1)^2} $

$ =\sqrt{16+36+144} $

$ =\sqrt{196}=14$

$ \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$

the shortest distance between the lines

$ =\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot\left(\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right)}{\left|\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right|}\right|$

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

$=\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{|2 \hat{i}-1 \hat{j}+0 \hat{k}|}\right|$

$\frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right|$

$ 3=|\lambda-4|$

$ \lambda-4= \pm 3 $

$\lambda=7,1$

Sum of all possible values of $\lambda$ is $=8$

$ \mathrm{M}(\lambda, 1+2 \lambda, 2+3 \lambda) $

$ \overrightarrow{\mathrm{PM}}=(\lambda-1) \hat{i}+(1+2 \lambda) \hat{j}+(3 \lambda-5) \hat{k}$
$\overrightarrow{\mathrm{PM}}$ is perpendicular to line $\mathrm{L}_1$
$ \overrightarrow{\mathrm{PM}} \overrightarrow{\mathrm{b}}=0 \quad(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) $
$ \Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0 $
$ 14 \lambda=14 \Rightarrow \lambda=1 $
$ \therefore \mathrm{M}=(1,3,5) $
$ \overrightarrow{\mathrm{Q}}=2 \overrightarrow{\mathrm{M}}-\overrightarrow{\mathrm{P}}[\mathrm{M} \text { is midpoint of } \overrightarrow{\mathrm{P}}  \overrightarrow{\mathrm{Q}}] $
$ \overrightarrow{\mathrm{Q}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}-\hat{\mathrm{i}}-7 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{Q}}=\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} $
$ \therefore(\alpha, \beta, \gamma)=(1,6,3)$
Required line having direction cosine $(l, \mathrm{~m}, \mathrm{n})$
$ l^2+m^2+n^2=1 $
$ \Rightarrow l^2+\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2=1$
$l^2=\frac{1}{4}$
$\therefore l=\frac{1}{2}$ [Line make acute angle with $\mathrm{x}$-axis]
Equation of line passing through $(1,6,3)$ will be
$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mu\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}\right)$
Option $(3)$ satisfying for $\mu=4$

$D.R$'s of $PQ$ are $2, 1,2$

$ \frac{2 s-2-t}{2}=\frac{s-t+2}{1}=\frac{s-t}{2} $

$ \Rightarrow t=6 \text { and } s=2 $

$ \Rightarrow P(6,4,6) \text { and } Q(2,2,2) $

$ P Q: \frac{x-2}{2}=\frac{y-2}{1}=\frac{z-2}{2}=\lambda$

Let $\mathrm{F}(2 \lambda+2, \lambda+2,2 \lambda+2)$

$\mathrm{A}(1,2,12)$

$\overrightarrow{\mathrm{AF}} \cdot \overrightarrow{\mathrm{PQ}}=0$

$\therefore \lambda=2$

So $\mathrm{F}(6,4,6)$ and $\mathrm{AF}=\sqrt{65}$

$M(4 \lambda+5, \lambda+4,3 \lambda+5) $

$L_2: \frac{\mathrm{x}+8}{12}=\frac{\mathrm{y}+2}{5}=\frac{\mathrm{z}+11}{9}=\mu $

$ N(12 \mu-8,5 \mu-2,9 \mu-11) $

$MN=(4 \lambda-12 \mu+13, \lambda-5 \mu+6,3 \lambda-9 \mu+16) . .(1)$

Now

$\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2=\left|\begin{array}{ccc}\hat{i} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 4 & 1 & 3 \\ 12 & 5 & 9\end{array}\right|=-6 \hat{\mathrm{i}}+8 \hat{\mathrm{k}}$   $.........(2)$

Equation $(1)$ and $(2)$

$\therefore \frac{4 \lambda-12 \mu+13}{-6}=\frac{\lambda-5 \mu+6}{0}=\frac{3 \lambda-9 \mu+16}{8}$

$I$ and $II$

$\lambda-5 \mu+6=0$  $..............(3)$

$I$ and $III$

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

Solve $(3)$ and $(4)$ we get

$ \lambda=-1, \mu=1 $

$ \therefore \mathrm{M}(1,3,2)$

$ \mathrm{N}(4,3,-2) $

$ \therefore \overrightarrow{\mathrm{OM}} \cdot \overrightarrow{\mathrm{ON}}=4+9-4=9 $

DR's $\rightarrow$ AP: $5 \mathrm{k}-4,2 \mathrm{k}-1,3 \mathrm{k}-7$

DR's $\rightarrow$ Line: $5,2,3$

Condition of perpendicular lines $(25 k-20)+(4 k-2)+(9 k-21)=0$

Then $\mathrm{k}=\frac{43}{38}$

Then $19(\alpha+\beta+\gamma)=101$

$\mathrm{d}_1=$ shortest distance between $\mathrm{L}_1 \& \mathrm{~L}_2$

$=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}\right|$

$ \mathrm{d}_1=2 $

$\mathrm{~L}_3: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+8}{-7}=\frac{\mathrm{z}-4}{5}, \mathrm{~L}_4: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-6}{-3} $

$ \mathrm{~d}_2=\text { shortest distance between } \mathrm{L}_3 \& \mathrm{~L}_4 $

$ \mathrm{~d}_2=\frac{12}{\sqrt{3}} \text { Hence } $

$ =\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}=16$

$\begin{aligned} & \mathrm{L}_1 \perp \mathrm{L}_2 \\ & 3-1+2 \mathrm{P}=0 \\ & \mathrm{P}=-1 \\ & \left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 2 \\ 3 & 1 & -1\end{array}\right|=-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\end{aligned}$

$\therefore(-\delta, 7 \delta, 4 \delta)$ will lie on $L_3$

For $\delta=1$ the point will be $(-1,7,4)$

$\mathrm{~N}(-3 \mu-2,-2 \mu+2,4 \mu+1) \quad \therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=-\mu+1$

$\frac{3 \lambda+2}{-3 \mu-1}=\frac{2 \lambda}{-2 \mu}=\frac{-2 \lambda-4}{4 \mu-2} $

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

$2 \mu=\lambda $

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

$\lambda \mu=\lambda+2 \mu \lambda \mu=2 \lambda$

$\Rightarrow \mu=2 \quad(\lambda \neq 0)$

$\therefore\lambda=4$

$ \alpha+\beta+\gamma=14$

$a+b+c=-1$

$\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}=196$

$\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 5 \\ -1 & 3 & 2\end{array}\right|$

Required line,

$\overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda^{\prime}(-9 \hat{\mathrm{i}}-9 \hat{\mathrm{j}}+9 \hat{\mathrm{k}})$

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

Now distance of $(0,2,-2)$

$\mathrm{P} . V . \text { of } \mathrm{P} \equiv(5+\lambda) \hat{\mathrm{i}}+(\lambda-4) \hat{\mathrm{j}}+(3-\lambda) \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{AP}}=(5+\lambda) \hat{\mathrm{i}}+(\lambda-6) \hat{\mathrm{j}}+(5-\lambda) \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{AP}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=0$

$5+\lambda+\lambda-6-5+\lambda=0 $

$\lambda=2$

$|\overrightarrow{\mathrm{AP}}|=\sqrt{49+16+9}$

$\mid \overrightarrow{\mathrm{AP}}=\sqrt{74}$

$\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1) $

$\overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k} $

$a=r+a-r=0 $

$2 a=2 r \rightarrow a=r $

$\overrightarrow{P R}=(a-k) i+(a+k) \hat{j}+(a+1) \hat{k} $

$a-k-a-k=0 \Rightarrow k=0 $

$A s, P Q \perp P R $

$(a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0 $

$a=1 \text { or }-1$

$12 a^2=12$

$(\lambda, 2,1) \&(\sqrt{3}, 1,2)$

$S.D$ $=\frac{\left|\begin{array}{ccc}\sqrt{3}-\lambda & -1 & 1 \\ -2 & 1 & 1 \\ 1 & -2 & 1\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1\end{array}\right|}$

$\begin{aligned} & 1=\left|\frac{\sqrt{3}-\lambda}{\sqrt{3}}\right| \\ & \lambda=0, \lambda=2 \sqrt{3}\end{aligned}$

$\overrightarrow{\mathrm{d}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\left(\mathrm{DR}\right.$ 's of $\left.L_2\right)$

$\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right|$

$=0 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}(\mathrm{DR}$ 's of Line perpendicular to   $\mathrm{L}_1$ and $\mathrm{L}_2$ )

$\mathrm{DR}$ of $\mathrm{AB}$ line

$=(0,2,2)=(3+\mu-\lambda, 3+\mu+\lambda, 3-\mu-\lambda)$

$\frac{3+\mu-\lambda}{0}=\frac{3+\mu+\lambda}{2}=\frac{3-\mu-\lambda}{2}$

Solving above equation we get $\mu=-\frac{3}{2}$ and $\lambda=\frac{3}{2}$

{ point }${A}=\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right)$

$\mathrm{B}=\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right)$

$\begin{aligned} & \text { Point of } \mathrm{AB}=\left(\frac{5}{2}, 2,6\right)=(\alpha, \beta, \gamma) \\ & 2(\alpha+\beta+\gamma)=5+4+12=21\end{aligned}$

$ \mathrm{PR}=6 $

$ (8 \lambda-4)^2+(2 \lambda+2)^2+(2 \lambda-4)^2=36 $

$ \lambda=0,1 $

$ \text { Hence } \mathrm{P}(-3,4,-1) \& \mathrm{Q}(5,6,1) $

$ \text { Centroid of } \Delta \mathrm{PQR}=(1,4,1) \equiv(\alpha, \beta, \gamma) $

$ \alpha^2+\beta^2+\gamma^2=18$

$ 3(3 \lambda-2)+2(2 \lambda-5)+(\lambda-7)=0 $

$14 \lambda=23 \Rightarrow \lambda=\frac{23}{14} $

$ \mathrm{~N}\left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}\right) $

$ \therefore \frac{\alpha+3}{2}=\frac{83}{14} \Rightarrow \alpha=\frac{62}{7} $

$ \frac{\beta+4}{2}=\frac{32}{14} \Rightarrow \beta=\frac{4}{7} $

$ \frac{\gamma+9}{2}=\frac{51}{14} \Rightarrow \gamma=\frac{-12}{7} $

$ \text { Ans. } 14(\alpha+\beta+\mathrm{r})=108$

$ \frac{38}{3 \sqrt{5}} \hat{\mathrm{k}}=\frac{19}{\sqrt{5}} $

$ \mathrm{k}=\frac{19}{\sqrt{5}} $

$ \mathrm{k}=\frac{3}{2} $

$ \int_0^{3 / 2}\left[\mathrm{x}^2\right]=\int_0^1 0+\int_1^{\sqrt{2}} 1+\int_{\sqrt{2}}^{3 / 2} 2 $

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

$ =2-\sqrt{2} $

$ \alpha=2 $

$ \Rightarrow 6 \alpha^3=48$

$DR'$ s of $A B \equiv \alpha-2: \beta-2: \gamma-2$

$\mathrm{AB} \perp_{\mathrm{ar}} \mathrm{L} \Rightarrow 2-\alpha+\beta-2+2 \gamma-4=0$

$2 \gamma+\beta-\alpha=4$

Let $C$ is mid-point of $A B$

$\mathrm{C}\left(\frac{\alpha+2}{2}, \frac{\beta+2}{2}, \frac{\gamma+2}{2}\right)$

$DR'$ s of PC $=\frac{\alpha}{2}: \frac{\beta-2}{2}: \frac{\gamma}{2}$

line $\mathrm{L} \| \mathrm{PC} \Rightarrow \frac{-\alpha}{2}=\frac{\beta-2}{2}=\frac{\gamma}{4}=\mathrm{K}$ (let)

$ \alpha=-2 \mathrm{~K} $

$ \beta=2 \mathrm{~K}+2 $

$ \gamma=4 \mathrm{~K}$

use in ($1$) $\Rightarrow \mathrm{K}=\frac{1}{6}$

value of $\alpha+\beta+6 \gamma=24 \mathrm{~K}+2=6$

$ \overrightarrow{\mathrm{AM}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})=0 $

$ (2 \lambda-7)(2)+(3 \lambda-6)(3)+(5 \lambda-5)(5)=0 $

$ 38 \lambda=57 $

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

$ \mathrm{M}\left(4, \frac{7}{2}, \frac{19}{2}\right) $

$ \mathrm{A}^{\prime}(0,2,12)$

$ \overrightarrow{\mathrm{RQ}}=\hat{\ell}-8 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{RS}}=\hat{\ell}+\hat{\mathrm{j}}-\hat{\mathrm{k}} $

$ \cos \theta=\frac{\overrightarrow{\mathrm{RQ}} \cdot \overrightarrow{\mathrm{RS}}}{|\overrightarrow{\mathrm{RQ}}||\overrightarrow{\mathrm{RS}}|}=\left|\frac{1-8-2}{\sqrt{69} \sqrt{3}}\right|=\frac{9}{3 \sqrt{23}} $

$ \cos \theta=\frac{3}{\sqrt{23}}=\frac{\mathrm{RS}}{\mathrm{RQ}}=\frac{\mathrm{RS}}{\sqrt{69}} $

$ \mathrm{RS}=3 \sqrt{3} $

$ \sin \theta=\frac{\sqrt{14}}{\sqrt{23}}=\frac{\mathrm{QS}}{\sqrt{69}} $

$ \mathrm{QS}=\sqrt{42} $

$ \operatorname{area}=\frac{1}{2} \cdot 2 \mathrm{QS} \cdot \mathrm{RS}=\sqrt{42} \cdot 3 \sqrt{3} $

$ \lambda=9 \sqrt{14} $

$ \lambda^2=81.14=14 \mathrm{k} $

$ \mathrm{k}=81$

$ \overline{\mathrm{a}}_2=-2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{p}}-=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} $

$ \overrightarrow{\mathrm{q}}-=-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} $

$(\lambda+2) \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}=\overline{\mathrm{a}}_1-\overline{\mathrm{a}}_2$

$ \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}-=-6 \hat{\mathrm{i}}-15 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$ \frac{44}{\sqrt{30}}=\frac{|-6 \lambda-12-105-9|}{\sqrt{(-6)^2+(-15)^2+3^2}} $

$ \frac{44}{\sqrt{30}}=\frac{|6 \lambda+126|}{3 \sqrt{30}} $

$ 132=|6 \lambda+126| $

$ \lambda=1, \lambda=-43 $

$ |\lambda|=43$

$=\left|\frac{(0 \hat{i}+2 \hat{j}+2 \hat{k}) \cdot(-15 \hat{i}+7 \hat{j}+9 \hat{k})}{\sqrt{355}}\right|$

$=\frac{0+14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}}$

$\therefore \mathrm{m}+\mathrm{n}=32+355=387$

 Shortest dist. $=\frac{\left|\overline{\mathrm{b}} \times\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right)\right|}{|\mathrm{b}|}=\frac{13}{\sqrt{29}} $

$ \frac{|(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times((2-\lambda) \hat{\mathrm{i}}+4 \hat{\mathrm{k}})|}{\sqrt{29}}=\frac{13}{\sqrt{29}} $

$ |-8 \hat{\mathrm{j}}-3(2-\lambda) \hat{\mathrm{k}}+12 \hat{\mathrm{i}}+4(2-\lambda) \hat{\mathrm{j}}|=13 $

$ |12 \hat{\mathrm{i}}-4 \lambda \hat{\mathrm{j}}+(3 \lambda-6) \hat{\mathrm{k}}|=13 $

$ 144+16 \lambda^2+(3 \lambda-6)^2=169 $

$ 16 \lambda^2+(3 \lambda-6)^2=25=\lambda \Rightarrow=1$

$\mathrm{L}_1: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\lambda$

$\mathrm{L}_2: \frac{\mathrm{x}+1}{\frac{1}{2}}=\frac{\mathrm{y}-1}{\frac{1}{2}}=\frac{\mathrm{z}+1}{1}=\mu$

dr of line $MN$ will be $<3+\lambda-\frac{\mu}{2},-1-\lambda-\frac{\mu}{2}, 2+\lambda-\mu>\&$ it will be proportional to $<3,1,2>$

$Image$

$\Rightarrow \lambda=-\frac{4}{3} \& \mu=-\frac{2}{3}$

$\therefore$ Coordinate of $\mathrm{M}$ will be $<\left(\frac{2}{3}, \frac{4}{3},-\frac{1}{3}\right)$ and equation of required line will be.

$\frac{\mathrm{x}-\frac{2}{3}}{3}=\frac{\mathrm{y}-\frac{4}{3}}{1}=\frac{\mathrm{z}+\frac{1}{3}}{2}=\mathrm{k}$

So any point on this line will be

$ \left(\frac{2}{3}+3 \mathrm{k}, \frac{4}{3}+\mathrm{k},-\frac{1}{3}+2 \mathrm{k}\right) $

$ \because \frac{2}{3}+3 \mathrm{k}=-\frac{1}{3} \Rightarrow \mathrm{k}=-\frac{1}{3}$

$\therefore$ Point lie on the line for

$\mathrm{k}=-\frac{1}{3} \text { is }\left(-\frac{1}{3}, 1,-1\right)$

$ \Rightarrow \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\lambda$

$Image$

$ \mathrm{B}(1+\lambda, 2+\lambda, 3+2 \lambda) $

$ \text { D.R. of } \mathrm{AB}=<\frac{3 \lambda-8}{3}, \frac{3 \lambda-5}{3}, \frac{6 \lambda-10}{3}> $

$ \mathrm{B}\left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \frac{3 \lambda-8}{3 \lambda-5}=\frac{2}{1} \Rightarrow 3 \lambda-8=6 \lambda-10 $

$ 3 \lambda=2 $

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

$ \mathrm{AB}=\frac{\sqrt{36+9+36}}{3}=\frac{9}{3}=3$

D.R of $\operatorname{PM}(3 \lambda-7,3 \lambda-4,5-\lambda)$

Since $P M$ is perpendicular to line

$\Rightarrow 3(3 \lambda-7)+3(3 \lambda-4)-1(5-\lambda)=0$

$\Rightarrow \lambda=2$

$\Rightarrow M (3,8,1) \Rightarrow PM =\sqrt{14}$

Let $Q=(\mu+2,2 \mu+2,3 \mu+3)$

$\Rightarrow \frac{2 \lambda-\mu-1}{1}=\frac{\lambda-2 \mu+1}{-1}=\frac{2 \lambda-3 \mu-1}{-2}$

$\Rightarrow \lambda=\mu=3 \Rightarrow P(7,6,8) \text { and } Q (5,8,12)$

$PQ =2 \sqrt{6}$

$\left|\frac{1}{2} \cdot 2 \sqrt{21} \cdot\right| \begin{array}{ccc} i & j & k \\ \alpha & 1 & \alpha+4 \\ 5 & 2 & 3\end{array}\left|\frac{1}{\sqrt{25+4+9}}\right|=21 \sqrt{21}$

$\sqrt{(2 \alpha+5)^2+(2 \alpha+20)^2+(2 \alpha-5)^2}=\sqrt{21} \sqrt{38}$

$\Rightarrow 12 \alpha^2+80 \alpha+450=798$

$\Rightarrow 12 \alpha^2+80 \alpha-348=0$

$\Rightarrow \alpha=3 \Rightarrow \alpha^2=9$

$\overline{ a }=(4,-2,-3)$

$\overline{ b }=(1,3,4)$

$\overline{ n }_1=(4,5,3)$

$\overline{ n }_2=(3,4,2)$

$\overline{ n }_1 \times \overline{ n }_2=\left|\begin{array}{lll} i & j & k \\ 4 & 5 & 3 \\ 3 & 4 & 2\end{array}\right|=\hat{ i }(-2)-\hat{ j }(-1)+\hat{ k }(1)=(-2,1,1)$

$S_d=\frac{(3,-5,-7) \cdot(-2,1,1)}{\sqrt{6}}=\left|\frac{-6-5-7}{\sqrt{6}}\right|=3 \sqrt{6}$

$\left|\begin{array}{lll}x_2-x_1 & a_1 & a_2 \\ y_2-y_1 & b_1 & b_2 \\ z_2-z_1 & c_1 & c_2\end{array}\right|=0$

Where $a_1, b_1, c_1$ are direction cosine of $1^{\text {st }}$ line and $a_2, b_2, c_2$ are direction cosine of $2^{\text {nd }}$ line.

Now, solving options

Point $(-3,1,5)$ and point $(-1,2,5)$

(1) $\left|\begin{array}{ccc}-3 & 1 & 5 \\ 1 & 2 & 5 \\ -2 & -1 & 0\end{array}\right|$

$=-3(5)-(10)+5(-1+4)$

$=-15-10+15=-10$

(2) Point $(-1,2,5)$

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

$=3(5)-(10)+5(1+4)$

$-25+25=0$

(3) Point $(-1,2,5)$

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

$-3(4)-(8)+5(1+4)$

$-12-8+25=5$

(4) Point $(-1,2,5)$

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

$-3(-5)-(-20)+5(-1-8)$

$15+20-45=-10$

$13=\frac{|153+8 \lambda|}{|4 \hat{ i }+3 \hat{ j }-12 \hat{ k }|}$

$=\frac{|153+8 \lambda|}{13}$
$|153+8 \lambda|=169$

$153+8 \lambda=169,-169$

$\lambda=\frac{16}{8}, \frac{-322}{8}$

$8\left|\sum_{\lambda \in S} \lambda\right|=306$

$\Rightarrow$ Shortest distance $=\frac{(\overrightarrow{ b }-\overrightarrow{ a }) \cdot(\overrightarrow{ p } \times \overrightarrow{ q })}{|\overrightarrow{ p } \times \overrightarrow{ q }|}$

S.D. $=(-\hat{ i }+2 \hat{ j }-\hat{ k }) \cdot \frac{(\overrightarrow{ p } \times \overrightarrow{ q })}{|\overrightarrow{ p } \times \overrightarrow{ q }|}$

$\left\{\overrightarrow{ p } \times \overrightarrow{ q } \equiv\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & \frac{1}{2} & \frac{-1}{12} \\ 1 & 1 & \frac{1}{6}\end{array}\right|=\frac{1}{6} \hat{ i }-\frac{1}{4} \hat{ j }+\frac{1}{2} \hat{ k }\right.$ or $\left.2 \hat{ i }-3 \hat{ j }+6 \hat{ k }\right\}$

S.D. $=\frac{(-\hat{ i }+2 \hat{ j }-\hat{ k }) \cdot(2 \hat{ i }-3 \hat{ j }+6 \hat{ k })}{\sqrt{2^2+3^2+6^2}}=\left|\frac{-14}{7}\right|=2$

$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3} \quad \vec{b}=\hat{i}+2 \hat{j}+6 \hat{k}$

$\vec{p}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \overrightarrow{ q }=2 \hat{i}+\hat{j}-3 \hat{k}$

$\overrightarrow{ p } \times \overrightarrow{ q }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & -7 & 5 \\ 2 & 1 & -3\end{array}\right|$

$=\hat{i}(16)-\hat{j}(-16)+\hat{k}(16)$

$=16(\hat{i}+\hat{j}+\hat{k})$

$d =\left|\frac{( a - b ) \cdot(\overrightarrow{ p } \times \overrightarrow{ q })}{|\overrightarrow{ p } \times \overrightarrow{ q }|}\right|=\left|\frac{(-10 \hat{ j }-2 \hat{ k }) \cdot 16(\hat{ i }+\hat{ j }+\hat{ k })}{16 \sqrt{3}}\right|$

$=\left|\frac{-12}{\sqrt{3}}\right|=4 \sqrt{3}$

$\frac{x-x_1}{a_1}=\frac{y-y_1}{a_2}=\frac{z-z_1}{a_3}$

$\frac{x-x_2}{b_1}=\frac{y-y_2}{b_2}=\frac{z-z_2}{b_3}$ is given as

$\frac{\left|\begin{array}{ccc}x_1-x_2 & y_1-y_2 & z_1-z_2 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3\end{array}\right|}{\sqrt{\left(a_1 b_3-a_3 b_2\right)^2+\left(a_1 b_3-a_3 b_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}$

$\frac{\left|\begin{array}{ccc}5-(3) & 2-(-5) & 4-1 \\ 1 & 2 & -3 \\ 1 & 4 & -5\end{array}\right|}{\sqrt{(-10+12)^2+(-5+3)^2+(4-2)^2}}$

$=\frac{\left|\begin{array}{ccc}8 & 7 & 3 \\ 1 & 2 & -3 \\ 1 & 4 & -5\end{array}\right|}{\sqrt{(2)^2+(2)^2+(2)^2}}$

$=\frac{16+14+6}{\sqrt{12}}=\frac{36}{\sqrt{12}}=\frac{36}{2 \sqrt{3}}$

$=\frac{18}{\sqrt{3}}=6 \sqrt{3}$

$\begin{array}{l}=\frac{\left|\begin{array}{ccc}4 & 2 & -14 \\3 & 2 & 2 \\3 & -2 & 0\end{array}\right|}{\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\3 & 2 & 2 \\3 & -2 & 0\end{array}\right|} \mid \\=\frac{16+12+168}{|-4 \hat{ i }+6 \hat{ j }-12 \hat{ k }|}=\frac{196}{14}=14\end{array}$

$\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$

$\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5}$ is $6$

Vector along line of shortest distance

$=\left|\begin{array}{lll} i & j & k \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|, \Rightarrow-\hat{ i }+2 \hat{ j }- k$ (its magnitude is $\sqrt{6}$ )

$\begin{array}{l}\text { Now } \frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\sqrt{6}+\lambda & \sqrt{6} & -3 \sqrt{6} \\2 & 3 & 4 \\3 & 4 & 5\end{array}\right|=\pm 6 \\\Rightarrow \lambda=-2 \sqrt{6}, 10 \sqrt{6}\end{array}$

So, square of sum of these values is $384$.

Let foot of perpendicular is

$P (2 \lambda-1,5 \lambda+1,-\lambda-1)$

$\overline{ PA }=(3-2 \lambda) \hat{ i }-(5 \lambda+1) \hat{ j }+(6+\lambda) \hat{ k }$

Direction ratio of line $\Rightarrow \vec{b}=2 \hat{i}+5 \hat{j}-\hat{k}$

$\text { Now, } \Rightarrow \overline{ PA } \cdot \overrightarrow{ b }=0$

$\Rightarrow 2(3-2 \lambda)-5(5 \lambda+1)-(6+\lambda)=0$

$\Rightarrow \lambda=\frac{-1}{6}$

$P (2 \lambda-1,5 \lambda+1,-\lambda-1) \equiv P (\alpha, \beta, \gamma)$

$\Rightarrow \alpha=2\left(-\frac{1}{6}\right)-1=-\frac{4}{3} \Rightarrow \alpha=-\frac{4}{3}$

$\Rightarrow \beta=5\left(-\frac{1}{6}\right)+1=\frac{1}{6} \Rightarrow \beta=\frac{1}{6}$

$\Rightarrow \gamma=-\lambda-1=\frac{1}{6}-1 \Rightarrow \gamma=-\frac{5}{6}$

$\therefore$ Check options

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

$\overrightarrow{ p } \times \overrightarrow{ q }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 1 & 1 \\ 2 & -1 & 0\end{array}\right|=\hat{ i }+2 \hat{ j }-3 \hat{ k }$

$d =\left|\frac{(\overrightarrow{ b }-\overrightarrow{ a }) \cdot(\overrightarrow{ p } \times \overrightarrow{ q })}{|\overrightarrow{ p } \times \overrightarrow{ q }|}\right|$

$d =\left|\frac{(-3 \hat{ j }-\hat{ k }) \cdot(\hat{ i }+2 \hat{ j }-3 \hat{ k })}{\sqrt{14}}\right|$

$=\left|\frac{-6+3}{\sqrt{14}}\right|=\frac{3}{\sqrt{14}}$

$\alpha=\frac{3}{\sqrt{14}}$

Now, $28 \alpha^2= 28 \times \frac{9}{14}=18$

$\therefore$ Equation of line is $\frac{ x -2}{1}=\frac{ y -3}{-1}=\frac{ z -1}{-1}$

Let $Q$ be $(5,3,8)$ and foot of $\perp$ from $Q$ on this line be $R$.

Now, $R \equiv( k +2,- k +3,- k +1)$

$D R$ of $Q R$ are $(k-3,-k,-k-7)$

$\therefore(1)( k -3)+(-1)(- k )+(-1)(- k -7)=0$

$\Rightarrow k =-\frac{4}{3}$

$\therefore \alpha^2=\left(\frac{13}{3}\right)^2+\left(\frac{4}{3}\right)^2+\left(\frac{17}{3}\right)^2=\frac{474}{9}$

$\therefore 3 \alpha^2=158$

$\begin{array}{l}\overline{ a _2}-\overline{ a _1}=6 \hat{ i }+(\lambda-1) \hat{ j }+(-\lambda-4) \hat{ k } \\ 2 \sqrt{6}=\left|\frac{-6-\lambda+1+2 \lambda+8}{\sqrt{1+1+4}}\right| \\ |\lambda+3|=12 \Rightarrow \lambda=9,-15 \\ \alpha=-2 k +5, \gamma= k -\lambda \text { where } k \in R \\ \Rightarrow \alpha+2 \gamma=5-2 \lambda=-13,35\end{array}$

$\begin{array}{ll}a_1=(0,0,0) & a_2=(3,0,5) \\b_1=(3,4,5) & b_2=(0,0,1)\end{array}$

Equation of edge parallel to $z$ axis

$\begin{aligned}& \frac{x-3}{0}=\frac{y-0}{0}=\frac{z-5}{1} \\& S . D=\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|} \\& \frac{\left|\begin{array}{lll}3 & 0 & 5 \\ 3 & 4 & 5 \\0 & 0 & 1\end{array}\right|}{\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{ k } \\3 & 4 & 5 \\0 & 0 & 1\end{array}\right|}=\frac{3(4)}{|4 \hat{i}-3 \hat{j}|}=\frac{12}{5} \\&\end{aligned}$

$\frac{\alpha-6}{3 / 2}=\frac{-1}{11-\beta}=\frac{2}{19} -19=22-2 \beta$

$\alpha-6=\frac{3}{19} 2 \beta=41$

$\alpha=6+\frac{3}{19}=\frac{117}{19}$

$\therefore(19 \alpha-6 \beta)^2=(117-123)^2=36$