Find the distance between two planes: C1:x+y+2z=4 and C2:3x+3y+6z=18 And find the other plane C3≠C1...

Well, we can rewrite ${\displaystyle C_2:3x+3y+6z=18}$
Two plain have normal ${\displaystyle n^{\to }=[3,3,6]}$
Chose ${\displaystyle P,Q}$ from ${\displaystyle C_1,C_2}$ respectively.
Say, ${\displaystyle P(5,0,0)}$ and ${\displaystyle Q(-\frac{1}{2},0,0)}$
Then, ${\displaystyle Q{P}^{\to }=[-\frac{11}{2},0,0]}$
${\displaystyle Proj\left\{{}_n^{\to }\right\}Q{P}^{\to }=\frac{Q{P}^{\to }{n}^{\to }}{{\left|\left|n^{\to }\right|\right|}^2}\times n^{\to }=\frac{-\frac{33}{2}}{q+q+36}[3,3,6]}$
${\displaystyle =-\frac{11}{36}[3,3,6]=[-\frac{11}{12},-\frac{11}{12},-\frac{11}{6}]}$
Thus, the distance is ${\displaystyle \left|\left|Proj\left\{{}_n^{\to }\right\}Q{P}^{\to }\right|\right|=\sqrt{{\left(\frac{11}{12}\right)}^2+{\left(\frac{11}{12}\right)}^2+{\left(\frac{11}{6}\right)}^2}=\sqrt{\frac{121}{24}}}$

For a plane defined by ${\displaystyle ax+by+cz=d}$ the normal is said to be (a,b,c). This is a direction, so we can normalise it ${\displaystyle \frac{(1,1,2)}{\sqrt{1+1+4}}=\frac{(3,3,6)}{\sqrt{9+9+36}}}$, which means these two planes are parallel and we can write the normal as ${\displaystyle \frac{1}{\sqrt{6}}(1,1,2)}$.
Now we should find two points on the planes.
Let ${\displaystyle y=0}$ and ${\displaystyle z=0}$, and find the corresponding ${\displaystyle x}$ values. For ${\displaystyle C_1,x=4}$ and for ${\displaystyle C_2,x=6}$. So we know ${\displaystyle C_1}$ contains the point ${\displaystyle (4,0,0)}$ and ${\displaystyle C_2}$ contains the point ${\displaystyle (6,0,0)}$.
The distance between these points is ${\displaystyle 2}$ and the direction is ${\displaystyle (1,0,0)}$. We now that this is not the shortest distance between these two points as ${\displaystyle (1,0,0)\ne \frac{1}{\sqrt{6}}(1,1,2)}$ so the direction is not perpendicular to these planes. But this is ok as we can use the dot product between ${\displaystyle (1,0,0)}$ and ${\displaystyle \frac{1}{\sqrt{6}}(1,1,2)}$ to work out the proportion of the distance that is perpendicular to the planes.
${\displaystyle (1,0,0)\times \frac{1}{\sqrt{6}}(1,1,2)=\frac{1}{\sqrt{6}}}$
So the distance between the two planes is ${\displaystyle \frac{2}{\sqrt{6}}}$.
The last part is to find the plane which is the same distance away from ${\displaystyle C_2}$ as ${\displaystyle C_1}$ but in the opposite direction. We know the normal must be the same, ${\displaystyle \frac{1}{\sqrt{6}}(1,1,2)}$. Using this we can write ${\displaystyle C_3:x+y+2z=a}$ and determine ${\displaystyle a}$. When ${\displaystyle y=0,z=0}$ we moved from ${\displaystyle (4,0,0)\to (6,0,0)}$, so if we move the same distance again we go ${\displaystyle (6,0,0)\to (8,0,0)}$ and ${\displaystyle (8,0,0)}$ is on ${\displaystyle C_3}$.
Therefore, ${\displaystyle a=8}$. So final equation of the plane is: ${\displaystyle C_3:x+y+2z=8}$

Thanks a lot!