Find the equation of the plane that passes through a point and parallel to another plane Find the plane given that it passes through the point $P = (0, -2, 5)$ and parallel to the plane $6x - y + 2z = 3$. 
If we're given $P$ in the equation is 
$$n \cdot r = n \cdot p$$
where $r = (x,y,z)$. However, if I take the vector from the general form and substitute it in for n, it doesn't make sense. <6,-1,2> Thanks!

So you have plane normal direction vector of $$\vec{n} = (6,-1,2)/|(6,-1,2)| =\frac{1}{\sqrt{46}} (6,-2,2)$$ and the distance of the plane to the origin $$d = 3/|\vec{n}|=\frac{3}{\sqrt{46}}$$
The new plane normal is identical to $\vec{n}$ and the new distance is
$$ r = \vec{n}\cdot\vec{p} = \frac{17}{\sqrt{46}}  $$
So your plane is
$$ \frac{1}{\sqrt{46}} (6,-1,3).(x,y,z) - r = 0 $$
  $$ 6x-y+3z-17 =0 $$
