JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Three Dimensional Geometry

A line equally inclined to the positive coordinate axes has direction $(1,1,1)$, so $L:\frac{x+3}{1}=\frac{y-5}{1}=\frac{z-2}{1}$ through $A(-3,5,2)$. For $P(-2,r,1)$, $\overrightarrow{AP}=(1,r-5,-1)$, and the distance of $P$ from $L$ is given by $d^2=|\overrightarrow{AP}{|}^2-\frac{(\overrightarrow{AP}\cdot \vec{d}{)}^2}{|\vec{d}{|}^2}=(2+(r-5{)}^2)-\frac{(r-5{)}^2}{3}=2+\frac{2}{3}(r-5{)}^2$. Setting $d^2=\frac{14}{3}$ gives $(r-5{)}^2=4$, so $r=3$ or $r=7$. The sum of all possible values of $r$ is $10$.