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

Divide the second equation by 2 to make coefficients of $x,y,z$ same:Plane 1: $2x+y+2z-8=0$ Plane 2: $2x+y+2z+2.5=0$ The distance $d$ between parallel planes $Ax+By+Cz+D_1=0$ and $Ax+By+Cz+D_2=0$ is:
$$d=\frac{|D_2-D_1|}{\sqrt{A^2+B^2+C^2}}$$
$$d=\frac{|2.5-(-8)|}{\sqrt{2^2+1^2+2^2}}=\frac{10.5}{\sqrt{9}}=\frac{10.5}{3}=3.5=\frac{7}{2}$$