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

\(2x+y+2z=8\)

\(2x+y+2z=-\frac{5}{2}\)

$D=\frac{|d_2-d_1|}{\sqrt{a^2+b^2+c^2}}$

$=\frac{\left|8+\frac{5}{2}\right|}{\sqrt{9}}=\frac{21}{6}=\frac{7}{2}$