JEE Main 2025 — 22 January, Morning Shift — Question 8

$\mathrm{P}(2 \lambda+1,3 \lambda+2,4 \lambda+3)$ on $\mathrm{L}_{1}$

$\mathrm{Q}(3 \mu+2,4 \mu+4,5 \mu+5)$ on $\mathrm{L}_{2}$

Dr's of $\mathrm{PQ}=3 \mu-2 \lambda+1,4 \mu-3 \lambda+2,5 \mu-4 \lambda+2$

$\mathrm{PQ} \perp \mathrm{L}_{1}$

$\Rightarrow(3 \mu-2 \lambda+1) 2+(4 \mu-3 \lambda+2) 3+(5 \mu-4 \underline{\lambda}+$ 2) $4=0$

$38 \mu-29 \lambda+16=0$ $\mathrm{PQ} \perp \mathrm{L}_{2}…(1)$

$\Rightarrow(3 \mu-2 \lambda+1) 3+(4 \mu-3 \lambda+2) 4+(5 \mu-4 \underline{\lambda}+$ 2) $5=0$

$50 \mu-38 \lambda+21=0…(2)$

By (1) & (2)

$\lambda=\frac{1}{3} ; \mu=\frac{-1}{6}$

$\therefore \mathrm{P}\left(\frac{5}{3}, 3, \frac{13}{3}\right) \& \mathrm{Q}\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)$

Line PQ $\frac{x-\frac{5}{3}}{\frac{1}{6}}$

$\frac{y-3}{\frac{-1}{3}} \quad \frac{z-\frac{13}{3}}{\frac{1}{6}}$

$\frac{x-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{z-\frac{13}{3}}{1}$

Point $\left(\frac{14}{3},-3, \frac{22}{3}\right)$ lies on the line PQ