The normal vector to $\alpha$ is ${\overrightarrow{n}}_{\alpha }=\left(1,-3,2\right)$, and the normal vector to $\beta$ is ${\overrightarrow{n}}_{\beta }=\left(1,-3,2\right)$. These two normal vectors are equal, which means they are parallel. Then we can find the distance between a point in $\beta$ and the plane $\alpha$.

1.

You find a point in the plane $\beta$ by letting two of the variables be $0$. Here, we set $y=0$ and $z=0$. Insert this into the equation of $\beta$, which will give you that $x=16$. That means $P=\left(0,0,16\right)$ is a point in the plane $\beta$. We already know that the normal vector to $\alpha$ is ${\overrightarrow{n}}_{\alpha }=\left(1,-3,2\right)$.

2.

Insert this information into the formula to find the distance between $\alpha$ and $\beta$: $$\begin{array}{llll}\hfill D& =\frac{\left|1\cdot 0-3\cdot 0+2\cdot 16-9\right|}{\sqrt{1^2+{\left(-3\right)}^2+2^2}}& \hfill & \\ \hfill & =\frac{\left|0+0+32-9\right|}{\sqrt{1+9+4}}& \hfill & \\ \hfill & =\frac{\left|23\right|}{\sqrt{14}}& \hfill & \\ \hfill & =\frac{23}{\sqrt{14}}.& \hfill & \end{array}$$