Distance between two parallel planes $2x+y+2z=8$ and $4x+2y+4z+5=0$ is

First, express both planes in the same form $Ax+By+Cz+D=0$.
Plane 1: $2x+y+2z-8=0$.
Plane 2: $4x+2y+4z+5=0\Rightarrow 2x+y+2z+5/2=0$.
The distance $d$ between parallel planes $Ax+By+Cz+D_1=0$ and $Ax+By+Cz+D_2=0$ is:
$$d=\frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$$
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$$d=\frac{|-8-5/2|}{\sqrt{2^2+1^2+2^2}}=\frac{|-21/2|}{\sqrt{9}}=\frac{21/2}{3}=7/2$$
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