JEE Main 2022 (Online) 25th July Morning ShiftQuestion 1
Let $P$ be the plane containing the straight line $\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$ and $\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$. If $d$ is the distance of $P$ from the point $(2,-5,11)$, then $d^2$ is equal to :
3D Geometry
Solution
Normal of the plane containing the two lines through O: $(2,3,5)\times(3,7,8)=(-11,-1,5)$. $P$ contains direction $(9,-1,-5)$ and is perpendicular to that plane, so its normal is $(-11,-1,5)\times(9,-1,-5)=(10,-10,20)\parallel(1,-1,2)$ (check: $9+1-10=0$). $P$ passes through $(3,-4,7)$: $x-y+2z=21$. $d=\frac{|2+5+22-21|}{\sqrt6}=\frac8{\sqrt6}$, so $d^2=\frac{64}{6}=\frac{32}3$. Option (C).