JEE 2024 — Mathematics PYQ

Let $A=(7,-2,11)$ and $B$ be any point on the line  

$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$  \hfill ...(i)

Now, let  

$\frac{x-7}{2}=\frac{y+2}{-3}=\frac{z-11}{6}=\lambda$  (say)

$x=2\lambda +7,y=-3\lambda -2,z=6\lambda +11$

$\therefore B=(2\lambda +7,-3\lambda -2,6\lambda +11)$

As point $B$ lies on (i), so  

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

$\Rightarrow -3\lambda -6=0\Rightarrow -3\lambda =6\Rightarrow \lambda =-2$

$\therefore B=(3,4,-1)$

Hence,  

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

$=\sqrt{16+36+144}=\sqrt{196}=14\text{ units}$