JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Three Dimensional Geometry

The plane $P$ contains the given line, so its normal is perpendicular to that line's direction $(9,-1,-5)$. The two lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$ and $\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$ span a plane whose normal is $(2,3,5)\times (3,7,8)=(-11,-1,5)$, and since $P$ is perpendicular to that plane, the normal of $P$ is also perpendicular to $(-11,-1,5)$. The normal of $P$ is therefore $(9,-1,-5)\times (-11,-1,5)=(-10,10,-20)$, proportional to $(1,-1,2)$. Passing through $(3,-4,7)$, the plane is $x-y+2z=21$. The distance of $(2,-5,11)$ from it is $\frac{|2+5+22-21|}{\sqrt{1+1+4}}=\frac{8}{\sqrt{6}}$, so $d^2=\frac{64}{6}=\frac{32}{3}$.