JEE Main 2025 — 7 April, Morning Shift — Question 38

$L: \frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c}$

$L_{1}: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ (say)

Any point on $L_{1}$ be $A(2 \lambda+1,3 \lambda-1,4 \lambda+1)$

$L_{2}: \frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}=\mu$ (say)

Any point on $L_{2}$ be $B(\mu+3,2 \mu-4, \mu)$

$D$ of $L$ be : $<2 \lambda, 3 \lambda-2,4 \lambda>$ or $<\mu+2,2 \mu+3$, $\mu-1>$

Now $\frac{2 \lambda}{\mu+2}=\frac{3 \lambda-2}{2 \mu+3}=\frac{4 \lambda}{\mu-1}$

$\Rightarrow \lambda=\frac{-6}{5} ,\quad \mu=-5$

$\therefore\langle a, b, c\rangle \equiv\langle-3,-7,-6\rangle$ or $\langle 3,7,6\rangle$

$\therefore \quad L: \frac{x-1}{3}=\frac{y-1}{7}=\frac{z-1}{6}$

$(7,15,13)$ lies on the line.