Derivation of Formula For Distance Between Parallel Planes

If two planes are parallel, the coefficients in the Cartesian equation are the same, since the coefficients of the Cartesian equation are the components of the normal, which is the same for parallel planes.
Suppose we want to find the distance between the parallel planes
$ax+by+cz=d_1$
(1)
$ax+by+cz=d_1$
(2)
Suppose a normal connects points
$(x_1,y_1,z_1)$
in plane (1) to point
$(x_2,y_2,z_2)$
in plane (2). Then
$ax_1+by_1+cz_1=d_1$
(3)
$ax_2+by_2+cz_2=d_1$
(4)
(4)-(3) gives
$a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1)=d_2-d_1$
(5)
Let the equation of the normal from plane (1) to (2) be
$\mathbf{r}(t)= \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} + \begin{pmatrix}a\\b\\c\end{pmatrix} t$
.
Then
$\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}= \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} + \begin{pmatrix}a\\b\\c\end{pmatrix} t_0 \rightarrow \begin{pmatrix}x_2-x_1\\y_2-y_1\\z_2-z_1\end{pmatrix}= \begin{pmatrix}a\\b\\c\end{pmatrix} t_0$

Substitution into (5) gives
$a^2t_0+b^2t_0+c^2t_0=d_2-d_1 \rightarrow t_0=\frac{d_2-d_1}{a^2+b^2+c^2}$

$\begin{pmatrix}x_2-x_1\\y_2-y_1\\z_2-z_1\end{pmatrix}-= \begin{pmatrix}a\\b\\c\end{pmatrix} \frac{d_2-d_1}{a^2+b^2+c^2}$

The distance between the planes is then
$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}=\sqrt{a^2+b^2+c^2}\frac{d_2-d_1}{a^2+b^2+c^2}= \frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}}$