JEE Main 2024 — 5 April, Shift 1 — Question 1

$\frac{\mathrm{x}+6}{3}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}+1}{1}=\lambda$

$\mathrm{x}=3 \lambda-6, \mathrm{y}=2 \lambda, \mathrm{z}=\lambda-1$

$\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu$

$\mathrm{x}=4 \mu+7, \mathrm{y}=3 \mu+9, \mathrm{z}=2 \mu+4$

$3 \lambda-6=4 \mu+7 \Rightarrow 3 \lambda-4 \mu=13$ ..(1)

$2 \lambda=3 \mu+9 \Rightarrow 2 \lambda-3 \mu=9$ ..(2)

After solving (1),(2) :

$\mu=-1$ , $\lambda=3$

int. point $(3,6,2) ;(7,8,9)$

$\mathrm{d}^{2}=16+4+49=69$

$\mathrm{d}^{2}+6=69+6=75$