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(C) The equations of the given planes are $2x - 2y + z + 3 = 0$ and $4x - 4y + 2z + 5 = 0$. First,we normalize the second equation so that the coeffic

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$\frac{1}{6}$

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(C) The equations of the given planes are $2x - 2y + z + 3 = 0$ and $4x - 4y + 2z + 5 = 0$. First,we normalize the second equation so that the coefficients of $x, y, z$ are the same as in the first equation. Dividing the second equation by $2$,we get $2x - 2y + z + \frac{5}{2} = 0$. The distance $d$ between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$ is given by the formula $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$. Here,$A = 2, B = -2, C = 1, D_1 = 3$,and $D_2 = \frac{5}{2}$. Substituting these values,we get $d = \frac{|3 - \frac{5}{2}|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|\frac{6-5}{2}|}{\sqrt{4 + 4 + 1}} = \frac{\frac{1}{2}}{\sqrt{9}} = \frac{\frac{1}{2}}{3} = \frac{1}{6}$.

The distance between the parallel planes $2x - 2y + z + 3 = 0$ and $4x - 4y + 2z + 5 = 0$ is:

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