Question

Point, Plane: $(3,-2,1),2x-y+2z+3=0$.

Answer 1

We know that the distance between a point $p(x_1,y_1,z_1)$ and a plane $Ax+By+Cz=D$ is given by,
$d=\left|\frac{A{x}_1+B{y}_1+C{z}_1-D}{\sqrt{A^2+B^2+C^2}}\right|....\left(1\right)$
Thus distance of point $(3,-2,1)$ from the plane $2x-y+2z+3=0$ is
$d=\left|\frac{2\times 3-(-2)+2\times 1+3}{\sqrt{{\left(2\right)}^2+{(-1)}^2+{\left(2\right)}^2}}\right|=\frac{13}{3}=\frac{13}{3}$