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Solution

Equations of planes: $\pi_1: 2x + y + 2z = 8$ $\pi_2: 4x + 2y + 4z + 5 = 0$ Normalize the second plane by dividing by 2: $\pi_2: 2x + y + 2z + \dfrac{5}{2} = 0$ Distance between parallel planes $\dfrac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} = \dfrac{|8 - (-\tfrac{5}{2})|}{\sqrt{2^2 + 1^2 + 2^2}} = \dfrac{\tfrac{21}{2}}{\sqrt{9}} = \dfrac{3}{2}$ units.