JEE Main 2026 (Online) 21st January Evening ShiftQuestion 1
Let the line $L$ pass through the point $(-3,5,2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $(-2,r,1)$ is $\sqrt{\frac{14}{3}}$, then the sum of all possible values of $r$ is :
3D Geometry
Solution
Direction $(1,1,1)$. Distance$^2=|d|^2-(d\cdot\hat n)^2$ with $d=(1,r-5,-1)$: $2+(r-5)^2-\frac{(r-5)^2}{3}=\frac{14}{3}$, so $(r-5)^2=4$, $r=7,3$, sum $10$.