JEE Main 2025 Apr 7 Shift 1, Mathematics Q18: Lines in Space

Parametrise a general point on each line.

On line 1: $A=(1+2\lambda ,-1+3\lambda ,1+4\lambda )$.

On line 2: $B=(3+\mu ,4+2\mu ,\mu )$.

The desired line $L$ passes through $C=(1,1,1)$ and both $A$ and $B$, so $\overrightarrow{CA}$ and $\overrightarrow{CB}$ must be parallel.

$\overrightarrow{CA}=(2\lambda ,3\lambda -2,4\lambda )$

$\overrightarrow{CB}=(\mu +2,2\mu +3,\mu -1)$

Setting components proportional: $\frac{\mu +2}{2\lambda }=\frac{2\mu +3}{3\lambda -2}=\frac{\mu -1}{4\lambda }$.

From the first and third ratio (eliminating $\lambda$): $\frac{\mu +2}{2}=\frac{\mu -1}{4}\cdot \frac{\lambda \cdot 2}{\lambda }$; more directly, $(\mu +2)(4\lambda )=(\mu -1)(2\lambda )\Rightarrow 2(\mu +2)=\mu -1\Rightarrow \mu =-5$.

Then $\overrightarrow{CB}=(-3,-7,-6)$, giving direction ratios $(3,7,6)$ for line $L$.

Equation of $L$: $\frac{x-1}{3}=\frac{y-1}{7}=\frac{z-1}{6}$.

Test each option: Substitute each point.

• $(7,15,13)$: $\frac{6}{3}=2$, $\frac{14}{7}=2$, $\frac{12}{6}=2$ ✓

$\therefore$ Option (4).