Question

Point, Plane: $(0,0,0),3x-4y+12z=3$.

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 $(0,0,0)$ from the plane $3x-4y+12z=3$ is
$d=\left|\frac{3\times 0-4\times 0+12\times 0-3}{\sqrt{{\left(3\right)}^2+{(-4)}^2+{\left(12\right)}^2}}\right|=\frac{3}{\sqrt{169}}=\frac{3}{13}$