Find the distance between the planes  $2\mathrm{x}-\mathrm{y}+2\mathrm{z}=5$ and $5\mathrm{x}-2.5\mathrm{y}+5\mathrm{z}=20$.

The distance between two parallel planes of the form $\mathrm{ax}+\mathrm{by}+\mathrm{cz}={\mathrm{d}}_1,\mathrm{ax}+\mathrm{by}+\mathrm{cz}={\mathrm{d}}_2$ is

$\begin{array}{l}\frac{\left|{\mathrm{d}}_1-{\mathrm{d}}_2\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}\\ 5\mathrm{x}-2.5\mathrm{y}+5\mathrm{z}=20\end{array}$

Given planes are 

2x-y+2z=5 and

5x-2.5y+5z=20

$5x-\frac{5}{2}y+5z=20$
On taking $\frac{5}{2}$common on both sides
The second equation can be written as $2\mathrm{x}-\mathrm{y}+2\mathrm{z}=8$

⇒ the distance between the given planes $2\mathrm{x}-\mathrm{y}+2\mathrm{z}=5$ and $2\mathrm{x}-\mathrm{y}+2\mathrm{z}=8$ is
$\frac{|8-5|}{\sqrt{2^2+1^2+2^2}}=1$
therefore  the distance between the given planes $2\mathrm{x}-\mathrm{y}+2\mathrm{z}=5$ and $2\mathrm{x}-\mathrm{y}+2\mathrm{z}=8$ is 1