QJEEM24Mad139: JEE Main 2024 Mathematics - Class 12 JEEM24Mad139 Solution: The distance, of the point from the line along the line , is :

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

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

$\frac{2 \lambda+7-6}{1}=\frac{-3 \lambda-2-4}{0}=\frac{6 \lambda+11-8}{3}$

$-3 \lambda-6=0$

$\lambda=-2$

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

$\mathrm{AB}=\sqrt{(7-3)^2+(4+2)^2+(11+1)^2}$

$=\sqrt{16+36+144}$

$=\sqrt{196}=14$