Question

Find the distance of the point $(-6,0,0)$ from the plane $2x-3y+6y=2$

Answer 1

Given $P:2x-3y+6y=2$

Point $(-6,0,0)$

Perpendicular distance of a point $(x_1,y_1,z_1)$ from $ax+by+c=0$ is $d=\left(\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}\right)$

$d=\left(\frac{|2(-6)-3\left(0\right)+6\left(0.2\right)|}{\sqrt{2^2+{(-3)}^2+6^2}}\right)=2$