Show that the distance D between parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is D = |d1 - d2| / sqrt(a2 + b2 + c2).

Let the first plane be $P_1: ax + by + cz + d_1 = 0$.
Let the second plane be $P_2: ax + by + cz + d_2 = 0$.
Since the planes are parallel, their normal vectors are the same, $\vec{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$.

Let $P_0(x_0, y_0, z_0)$ be a point on the first plane $P_1$.
Therefore, $ax_0 + by_0 + cz_0 + d_1 = 0$.
From this equation, we can express $ax_0 + by_0 + cz_0 = -d_1$.

The distance $D$ from a point $(x_0, y_0, z_0)$ to a plane $Ax + By + Cz + D_{plane} = 0$ is given by the formula:

We want to find the distance from the point $P_0(x_0, y_0, z_0)$ (which lies on $P_1$) to the second plane $P_2: ax + by + cz + d_2 = 0$.
In this case, $A=a$, $B=b$, $C=c$, and $D_{plane}=d_2$.

Substituting the coordinates of $P_0$ and the coefficients of $P_2$ into the distance formula:

We know from Step 2 that $ax_0 + by_0 + cz_0 = -d_1$. Substitute this into the equation for $D$:

Since $|d_2 - d_1|$ is equivalent to $|d_1 - d_2|$, we can write:

This shows the required formula for the distance between two parallel planes.