JEE Main 2023 (Online) 31st January Morning ShiftQuestion 1
Let the shortest distance between the lines $L:\frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1}$, $\lambda\ge0$ and $L_1: x+1=y-1=4-z$ be $2\sqrt6$. If $(\alpha,\beta,\gamma)$ lies on $L$, then which of the following is NOT possible?
3D Geometry
Solution
Cross of directions $(-2,0,1)$, $(1,1,-1)$ is $(-1,-1,-2)$; distance $\frac{|3+\lambda|}{\sqrt6}=2\sqrt6$, so $\lambda=9$. Points on $L$: $(5-2t,9,-9+t)$. Then $\alpha+2\gamma=-13$ always, so $24$ is impossible.