The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)$ is:

(A) $2\sqrt 6$
(B) $\sqrt {26}$
(C) $2\sqrt 5$
(D) $4\sqrt 2$

Solution

We have, $3x-z+4=0$ or $z=3x+4$
& $3y-2z-1=0$ or $3y-2(3x+4)-1=0$ or $y=2x+3$

Any point $(x,y,z)$ on the line $\equiv (t,2t + 3,3t + 4)$

Let $d$ be the distance between $(2,-1,6)$ & $(t,2t + 3,3t + 4)$

Then, $d^2=(t-2)^2+(2t+3+1)^2+(3t+4-6)^2=14t^2+24$

Minimum $d$ = Required answer = $\sqrt {24}=2\sqrt 6$ when t = 0.

Answer: (A)