Find the distance between the parallel planes $x - y + 3z + 5 = 0$ $\;$ and $\;$ $2x - 2y + 6z + 7 = 0$
Equations of the given planes are
$x - y + 3z + 5 = 0$ $\;\;\; \cdots \; (1)$
$2x - 2y + 6z + 7 = 0$ $\implies$ $x - y + 3z + \dfrac{7}{2} = 0$ $\;\;\; \cdots \; (2)$
Equations $(1)$ and $(2)$ are of the form
$ax + by + cz + d_1 = 0$ $\;$ and $\;$ $ax + by + cz + d_2 = 0$ $\;$ respectively.
Here $\;$ $a = 1, \; b = -1, \; c = 3, \; d_1 = 5, \; d_2 = \dfrac{7}{2}$
Distance between the parallel planes $(1)$ and $(2)$ is
$\begin{aligned}
\left|\dfrac{d_1 - d_2}{\sqrt{a^2 + b^2 + c^2}}\right| & = \left|\dfrac{5 - \dfrac{7}{2}}{\sqrt{\left(1\right)^2 + \left(-1\right)^2 + \left(3\right)^2}}\right| \\\\
& = \left|\dfrac{3}{2 \sqrt{1 + 1 + 9}}\right| \\\\
& = \dfrac{3}{2 \sqrt{11}} \;\; units
\end{aligned}$