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

The given planes are

$$2x+y+2z-8=0$$

and

$$4x+2y+4z+5=0$$

Dividing the second equation by $$2$$,

$$2x+y+2z+\frac52=0$$

Now both planes have the same coefficients of

$$x,\ y,\ z$$

Hence they are parallel.

Using the formula for the distance between parallel planes

$$ax+by+cz+d_1=0$$

and

$$ax+by+cz+d_2=0$$

the distance is

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

Here,

$$d_1=-8,\qquad d_2=\frac52$$

Therefore,

$$D=\frac{\left|-8-\frac52\right|}{\sqrt{2^2+1^2+2^2}}$$

$$=\frac{\left|-\frac{21}{2}\right|}{3}$$

$$=\frac{21}{6}$$

$$=\frac{7}{2}$$