Chapter 11 - Section 11.5 - Lines and Planes in Space - Exercises - Page 631: 42

$\dfrac{8}{3}$

The formula to calculate the distance for two vectors is given by: $d=\dfrac{|u \cdot v|}{|v|}$ Thus, we have $u \cdot v=-2(2)+2(1)-3(2)=-8$ and $|u \times v|=|-8|=8$ Now, $d=\dfrac{|u \cdot v|}{|v|}=\dfrac{8}{ \sqrt {(2)^2+(1)^2+(2)^2}}=\dfrac{8}{ \sqrt {9}}=\dfrac{8}{3}$