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
This is a sort of geometric way to see this, suppose instead of the finding the distance between the planes $C_1$ and $C_2$, we consider the simpler problem of finding the distance between the planes
$z=2$ and $z=4$
It is clear, in this case that the distance between these planes is $2$ and is realized by the points $(0,0,2)$ and $(0,0,4)$ respectively. Notice, that these points are the intersection of the line parametrized by $(0,0,t)$ and the planes themselves. What's special about this line, is that it is the line paramaterized by the unit vector $[0,0,1]$.
Now, if we turn our heads to the current problem $C_{1}$ and $C_{2}$, we can essentially use the same argument by tilting our heads and treating the normal vector the planes $[1,1,2]$ as the $z$-axis above. Doing this, we see that the intersection of the line $(t,t,2t)$ and plane $C_{1}$ is $(\frac{2}{3}, \frac{2}{3}, \frac{4}{3})$ and the intersection with $C_{2}$ is $(1,1,2)$. Computing the distance between these two points we obtain $\frac{\sqrt{2}}{\sqrt{3}}$ or $\frac{2}{\sqrt{6}}$.
The reason why this works is that we are using $[1,1,2]$ and the plane $x+y+2z=0$ through the origin as a basis for $\mathbb{R}^{3}$. It's a change of coordinates of sorts.