JEE Main 2021 — Three Dimensional Geometry Question with Solution

$P\left(3,-1,2\right)$

$Q\left(1,2,-4\right)$

$\overrightarrow{PR} \Vert 4\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}$

$\overrightarrow{QS} \Vert -2\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}$

Dr's of normal to the plane containing $P,T$ and $Q$ will be proportional to : 

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

$\therefore \frac{\ell }{0}=\frac{m}{4}=\frac{n}{2}$

For point, $T$ : $\overrightarrow{PT}=\frac{x-3}{4}=\frac{y+1}{-1}=\frac{z-2}{2}=\lambda$

$\overrightarrow{QT}=\frac{x-1}{-2}=\frac{y-2}{1}=\frac{z+4}{-2}=\mu$

$T\equiv \left(4\lambda +3,-\lambda -1,2\lambda +2\right)$

$Q\equiv \left(2\mu +1,\mu +2,-2\mu -4\right)$

$4\lambda +3=-2\mu +1\Rightarrow 2\lambda +\mu =-1$

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

and $\mu =-5, \lambda +\mu =-3\Rightarrow \lambda =2$

So point $T:\left(11,-3,6\right)$

$\overrightarrow{\mathit{O}\mathit{A}}\mathit{=}\left(11\hat{\mathrm{i}}-3\hat{\mathrm{j}}+6\hat{\mathrm{k}}\right)\pm \left(\frac{2\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{5}}\right)\sqrt{5}$

$\overrightarrow{OA}=\left(11\hat{\mathrm{i}}-3\hat{\mathrm{j}}+6\hat{\mathrm{k}}\right)\pm \left(2\hat{\mathrm{j}}+\hat{\mathrm{k}}\right)$

$\overrightarrow{OA}=11\hat{\mathrm{i}}-\hat{\mathrm{j}}+7\hat{\mathrm{k}}$ or $9\hat{\mathrm{i}}-5\hat{\mathrm{j}}+5\hat{\mathrm{k}}$

$\left|\overrightarrow{OA}\right|=\sqrt{121+1+49}=\sqrt{171}$ or $\sqrt{81+25+25}=\sqrt{131}$