JEE 2023 — Mathematics PYQ

Since $\left(\frac{5}{2},1,\lambda \right)$ and $(-2,0,1)$ are equidistant from plane $2x+3y-6z+7=0$

$\Rightarrow \left|\frac{2\left(\frac{5}{2}\right)+3(1)-6(\lambda )+7}{\sqrt{2^2+3^2+6^2}}\right|=\left|\frac{2(-2)+3(0)-6(1)+7}{\sqrt{2^2+3^2+6^2}}\right|$

$\Rightarrow \left|\frac{5+3-6\lambda +7}{\sqrt{4+9+36}}\right|=\left|\frac{-4-6+7}{\sqrt{4+9+36}}\right|$

$\Rightarrow \left|\frac{15-6\lambda }{\sqrt{49}}\right|=\left|\frac{-3}{\sqrt{49}}\right|$

$\Rightarrow 15-6\lambda =\pm 3$

$\Rightarrow 15-6\lambda =3$ or $15-6\lambda =-3$

$\Rightarrow 6\lambda =12$ or $6\lambda =18$

$\Rightarrow {\lambda }_1=2$ or ${\lambda }_2=3$

So, point will be $(1,2,3)$

Let $\overrightarrow{M_0}=(1,2,3)$

$\overrightarrow{M_1}$ is the point through which line passes i.e., $(5,1,-7)$ and $\vec{s}=\hat{i}+2\hat{j}+2\hat{k}$

$\therefore \overrightarrow{M_0\overrightarrow{M_1}}=4\hat{i}-\hat{j}-10\hat{k}$

Now, required distance = $\frac{\left|\overrightarrow{M_0\overrightarrow{M_1}}\times \vec{s}\right|}{|\vec{s}|}$

= $\frac{|(4\hat{i}-\hat{j}-10\hat{k})\times (\hat{i}+2\hat{j}+2\hat{k})|}{\sqrt{1^2+4^2+4^2}}$

= $\frac{|18\hat{i}-18\hat{j}+9\hat{k}|}{3}$

= 9