JEE Main 2026 Jan 28 Shift 2, Mathematics Q16: Lines in Space

Find foot of perpendicular $N$ from $P(3,2,1)$ to the first line. Parametrise: $N=(1+r,2r,1+r)$. Direction ratios of $\overrightarrow{PN}=(1+r-3,2r-2,1+r-1)=(r-2,2r-2,r)$. Direction of line $=⟨1,2,1⟩$. Perpendicularity:

$1(r-2)+2(2r-2)+1(r)=0\Rightarrow 6r-6=0\Rightarrow r=1$.

So $N=(2,2,2)$. Image $Q$ satisfies $N=\frac{P+Q}{2}$, giving $Q=(1,2,3)$.

Distance from $Q(1,2,3)$ to the second line through $A=(9,9,5)$ with direction $\vec{d}=⟨3,2,-2⟩$:

$\overrightarrow{AQ}=⟨1-9,2-9,3-5⟩=⟨-8,-7,-2⟩$. $|\overrightarrow{AQ}{|}^2=64+49+4=117$.

Projection on $\vec{d}$: $\overrightarrow{AQ}\cdot \vec{d}=-24-14+4=-34$. $|\vec{d}|=\sqrt{17}$, so $|AM|=\frac{|-34|}{\sqrt{17}}=2\sqrt{17}$, and $|AM{|}^2=68$.

$QM=\sqrt{|AQ{|}^2-|AM{|}^2}=\sqrt{117-68}=\sqrt{49}=7$.