JEE Main 2025 Jan 22 Shift 1, Mathematics Q8: Lines in Space

Parametrise: $P=(2\lambda +1,3\lambda +2,4\lambda +3)$ on $L_1$ and $Q=(3\mu +2,4\mu +4,5\mu +5)$ on $L_2$.

Direction of $\overrightarrow{PQ}$ is $(3\mu -2\lambda +1,4\mu -3\lambda +2,5\mu -4\lambda +2)$. For the shortest distance segment, $\overrightarrow{PQ}$ must be perpendicular to both lines:

$\overrightarrow{PQ}\cdot (2,3,4)=0\Rightarrow 38\mu -29\lambda +16=0$ ... (1)

$\overrightarrow{PQ}\cdot (3,4,5)=0\Rightarrow 50\mu -38\lambda +21=0$ ... (2)

Solving: $\lambda =\frac{1}{3}$, $\mu =-\frac{1}{6}$.

$P=\left(\frac{5}{3},3,\frac{13}{3}\right)$ and $Q=\left(\frac{3}{2},\frac{10}{3},\frac{25}{6}\right)$. Direction $PQ$ simplifies to proportional to $(1,-2,1)$.

Line PQ: $\frac{x-5/3}{1}=\frac{y-3}{-2}=\frac{z-13/3}{1}$. Checking $\left(\frac{14}{3},-3,\frac{22}{3}\right)$: $\frac{14/3-5/3}{1}=3$; $\frac{-3-3}{-2}=3$; $\frac{22/3-13/3}{1}=3$. All equal, so the point lies on the line.