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

Parallel planes are level sets of a linear function. In this case, $x+y+2z=c$.
The signed distance of $x+y+2z=c$ to the origin is the normalized algebraic value
$$
\frac{c}{\sqrt{1^2+1^2+2^2}}=\frac{c}{\sqrt{6}}
$$
Therefore, the unsigned distance between two planes $x+y+2z=c_1$ and $x+y+2z=c_2$ is
$$
\frac{|c_1-c_2|}{\sqrt{6}}
$$
In your example, $c_1=4$ and $c_2=6$ and so the distance is
$$
\frac{2}{\sqrt{6}}
$$
The other plane at the same distance to $C_2$ has $c=6+2=8$ and so is given by $x+y+2z=8$.
