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(B) The equations of the two parallel planes are given by $r \cdot n + d_1 = 0$ and $r \cdot n + d_2 = 0$.Here,$n = i + 2j - 2k$,$d_1 = 5$,and $d_2 =

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$\frac{13}{3} 	ext{ units}$

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(B) The equations of the two parallel planes are given by $r \cdot n + d_1 = 0$ and $r \cdot n + d_2 = 0$.Here,$n = i + 2j - 2k$,$d_1 = 5$,and $d_2 = -8$.The distance $D$ between two parallel planes $r \cdot n + d_1 = 0$ and $r \cdot n + d_2 = 0$ is given by the formula $D = \frac{|d_1 - d_2|}{|n|}$.First,calculate the magnitude of the normal vector $n$:$|n| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.Now,substitute the values into the distance formula:$D = \frac{|5 - (-8)|}{3} = \frac{|5 + 8|}{3} = \frac{13}{3} 	ext{ units}$.

The distance between the planes given by $r \cdot (i + 2j - 2k) + 5 = 0$ and $r \cdot (i + 2j - 2k) - 8 = 0$ is

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