JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Three Dimensional Geometry

The first line is $\vec{r}=(1,0,1)+r(1,2,1)$; the foot of perpendicular from $P(3,2,1)$ is $N=(1+r,2r,1+r)$. Requiring $\overrightarrow{PN}=(r-2,2r-2,r)$ perpendicular to $(1,2,1)$ gives $(r-2)+2(2r-2)+r=0$, so $6r-6=0$, $r=1$ and $N=(2,2,2)$. The image is $Q=2N-P=(1,2,3)$. For the second line through $A(9,9,5)$ with direction $\vec{b}=(3,2,-2)$, take $\overrightarrow{AQ}=(-8,-7,-2)$. Then $\overrightarrow{AQ}\times \vec{b}=(18,-22,5)$, with $|\overrightarrow{AQ}\times \vec{b}|=\sqrt{324+484+25}=\sqrt{833}$ and $|\vec{b}|=\sqrt{17}$. The distance is

The distance of $Q$ from the line is $7$.