Point above a plane

The distance is the length of the projection of the point onto the normal of the plane.

1. The vector $n=(2,-1,2)$ is normal to the plane, and its length is $\sqrt{4+1+4}=3$.
2. The dot product of the point with $n$ is $2\times 1-1\times 2+2\times 3=6$.
3. The projection onto $n$ has length equal to that dot product divided by the length of $n$, so the distance is $\frac{6}{3}=2$.

Answer: $2$