Q:

State the possible Cartesian equations of $p_2$.

Given a plane $p_1$ to  have the equation $x+2y-3z=0$ with points $B = (0,3,2), M = (1.5,3,2.5)$. Another plane $p_2$ has a perpendicular distance of $0.5$ with $p_1$. State the possible Cartesian equations of $p_2$.
I think I used a wrong way but got the correct answer, I let $Q$ be a point $(x,y,z)$ on $p_2$, and I know $BQ = (x,y-3,z-2)$ then i use length of projection formula to get $$|x+2y-3z| = \frac{2}{\sqrt{14}}$$
Then I let $x = \frac{2}{\sqrt{14}}, y = 0, z=0$ to get a point $Q$.

A:

The plane $p_2$ is parallel to $p_1$ so it is of the form $x+2y-3z=c$.
The plane $p_1$ passes through the origin so the distance of $p_1$ and $p_2$ is equal to the distance of the origin and $p_2$.
Distance of a plane $Ax+By+Cz = D$ from the origin is given by $\frac{|D|}{\sqrt{A^2+B^2+C^2}}$ so in this case we have
$$0.5 = d(0, p_2) = \frac{|c|}{\sqrt{1^2+2^2+3^2}} = \frac{|c|}{\sqrt{14}}$$
so $c = \pm \frac{\sqrt{14}}2$. Therefore $p_2$ is given by 
$x+2y-3z=\pm \frac{\sqrt{14}}2$.

