Answer
$distance = \frac{4}{5}$
Work Step by Step
Since ${\bf{n}} = \left( {\frac{3}{5},\frac{4}{5},0} \right)$, the plane has equation:
${\bf{n}}\cdot\left( {x,y,z} \right) = 3$
$\left( {\frac{3}{5},\frac{4}{5},0} \right)\cdot\left( {x,y,z} \right) = 3$
$\frac{3}{5}x + \frac{4}{5}y = 3$
Using Eq. (8), the distance from $Q = \left( {1,2,2} \right)$ to the plane $\frac{3}{5}x + \frac{4}{5}y = 3$ is
$distance = \frac{{\left| {\frac{3}{5}\cdot1 + \frac{4}{5}\cdot2 - 3} \right|}}{{||\left( {\frac{3}{5},\frac{4}{5},0} \right)||}}$
$distance = \frac{{\left| {\frac{{11}}{5} - 3} \right|}}{{\sqrt {\frac{9}{{25}} + \frac{{16}}{{25}}} }} = \frac{4}{5}$
