JEE Main 2023 (Online) 15th April Morning ShiftQuestion 1
Let $S$ be the set of all values of $\lambda$, for which the shortest distance between the lines $\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$ is $13$. Then $8\left|\sum_{\lambda\in S}\lambda\right|$ is equal to
3D Geometry
Solution
Points $(\lambda,3,-6)$, $(-\lambda,0,6)$; directions $(0,4,1)$, $(3,-4,0)$. Cross $=(4,3,-12)$, $|n|=13$. Difference of points $=(-2\lambda,-3,12)$. Dot: $-8\lambda-9-144=-8\lambda-153$. $|8\lambda+153|/13=13\Rightarrow 8\lambda+153=\pm169\Rightarrow \lambda=2$ or $-\frac{161}{4}$. Sum $=-\frac{153}{4}$, so $8|\sum|=306$. Option (C).