Question

The distance of a point $(1,-2,3)$ from the plane $x-y+z=5$ and parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ is

• A. $7$
• B. $1$
• C. $3$
• D. $13$

Answer 1

The correct option is B $1$

let line equation parallel to given line is $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}$

Suppose this line meets the plane at a point $\left(2r+1,3r-2,-6r+3\right)$

So substituting this point in the plane equation

$2r+1-3r+2-6r+3=5$

$-7r+1=0$

$r=\frac{1}{7}$

$d=\sqrt{{\left(2r\right)}^2+{\left(3r\right)}^2+{\left(-6r\right)}^2}$

$=\sqrt{4+9+36r^2}$

$=7\times \frac{1}{7}=1$

Then,

Option $B$ is correct answer.