Planes in 3D

General plane equation

A x + B y + C z = D

Normal to a plane

\vec{n} = A \vec{i} + B \vec{j} + C \vec{k}

Plane by normal and point

Given $\vec{n}$ - normal and point $\vec{p}$ on the plane $A x + B y + C z = D$
?if I got it right?

n_{1} (x - p_{1}) + n_{2} (y - p_{2}) + n_{3} (z - p_{3}) = 0

Point distance to plane

Given a point $\vec{p}$ on the plane $A x + B y + C z = D$
the distance $d$ from the point to the plane will be:

d = \frac{A p_{1} + B p_{2} + C p_{3} - D}{\sqrt{A^{2} + B^{2} + C^{2}}}