Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 42

$\dfrac{8}{3}$

The distance formula for two vectors can be calculated as: $d=\dfrac{|p \cdot q|}{|q|}$ Now, $p \cdot q=-2(2)+2(1)-3(2)=-8$ and $|p \times q|=|-8|=8$ Thus, $d=\dfrac{8}{ \sqrt {(2)^2+(1)^2+(2)^2}}=\dfrac{8}{ \sqrt {4+1+4}}$ or, $=\dfrac{8}{ \sqrt {9}}$ or, $=\dfrac{8}{3}$