How to find the distance between two planes? The following show you the whole question.

Find the distance d bewteen two planes
  \begin{eqnarray}
\\C1:x+y+2z=4 \space \space~~~ \text{and}~~~ \space \space C2:3x+3y+6z=18.\\
\end{eqnarray}
  Find the other plane $C3\neq C1$ that has the distance d to the plane $C2$.

According to the example my teacher gave me, the answer should be :


Am I right? However, I do not know what is normal and why there are P(5) and Q($-\frac{1}{2}$).

Thank you for your attention

The term "normal" means perpendicularity. A normal vector of a plane in three-dimensional space points in the direction perpendicular to that plane. (This vector is unique up to non-zero multiplies.)
For determining the distance between the planes $C_1$ and $C_2$ you have to understand what is the projection of a vector onto the direction of a second one, see https://en.wikipedia.org/wiki/Vector_projection.
The the projection of any vector $\vec{QP}$ pointing from one of the planes to the other onto the direction of the common normal vector $\vec{n}=(1,1,2)$ (or $\vec{n}=(3,3,6)$ because length does not matter but only direction does) is a vector perpendicular to both planes and pointing from one to another. So its length has to be the distance.
Finding $C_3$ is easy as well. $C_3$ is the image of $C_1$ under reflection about $C_2$. So $C_3$ is the set of points $(x,y,z)$ which can be written as $2(x'',y'',z'')-(x',y',z')$ with $(x'',y'',z'')$ in $C_2$ and $(x',y',z')$ in $C_1$. Checking what is $3$(first coordinate)$+3$(second coordinate)$+6$(third coordinate) for the points of $C_3$, we get $3(2x''-x')+3(2y''-y')+6(2z''-z')=2(3x''+3y''+6z'')-3(x'+y'+z')=2\cdot18-3\cdot 4=12$. Hence the equation for the plane $C_3$ is $3x+3y+6z=24$ or $x+y+2z=8$. 
