JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Three Dimensional Geometry

Let $Q$ be the reflection of $P(a,2,a)$ in the first line, which passes through $(0,-a,0)$ with direction $(2,1,1)$. Reflecting $Q$ in the second line, of direction $(2,1,-5)$, must return $P$, so $\overrightarrow{PQ}$ is perpendicular to both directions and the midpoint of $PQ$ lies on both lines. Perpendicularity of $\overrightarrow{PQ}$ to $(2,1,-5)$ yields $4a-4=0$, so $a=1$; the foot on the first line is then the midpoint $M=(2,0,1)$. Requiring $M$ on the second line $\frac{x-2b}{2}=\frac{y-1}{1}=\frac{z+2b}{-5}$ gives $\frac{2-2b}{2}=-1$, so $b=2$. Therefore $a+b=3$.