JEE Main 2024 — 27 January, Shift 1 — Question 2

$\mathrm{B}=(2 \lambda+7,-3 \lambda-2,6 \lambda+11)$

Point B lies on $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ $\frac{2 \lambda+7-6}{1}=\frac{-3 \lambda-2-4}{0}=\frac{6 \lambda+11-8}{3}$

$-3 \lambda-6=0$

$\lambda=-2$

$\mathrm{B} \Rightarrow(3,4,-1)$

$$\begin{aligned}A B & =\sqrt{(7-3)^{2}+(4+2)^{2}+(11+1)^{2}} & =\sqrt{16+36+144} & =\sqrt{196}=14\end{aligned}$$