Question

The distance of the point $(3,8,2)$ from the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ measured parallel to the plane $3x+2y-2z=0$ is -

• A. $2$
• B. $3$
• C. $6$
• D. $7$

Answer 1

The correct option is D $7$

Let $(3,8,2)$ be denoted by $P$ and $Q$ be any point on the given line,

$Q(2r+1,4r+3,3r+2)$ and the direction ratio of the plane is $\overrightarrow{n}=(3,2,-2)$

$\Rightarrow \overrightarrow{PQ}=(2r-2,4r-5,3r)$

$\overrightarrow{PQ}\cdot \overrightarrow{n}=0\Rightarrow r=2\Rightarrow \overrightarrow{PQ}=(2,3,6)\Rightarrow |\overrightarrow{PQ}|=7$