Find a plane $\Pi'$ given two points, a plane, and a prescribed distance. I have to work this out:

Let $\Pi:2x-y+2z=4$, $P=(2,2,2)$ and $Q=(1,0,1)$. Find a plane $\Pi'$ containing $P$ and $Q$, and a point $R$ of $\Pi$ such that $d(P,R)=d(P,\Pi)$.

Now, observe that $Q$ is in $\Pi$, so we can find $d(P,\Pi)$ as follows:
$$\eqalign{
   d\left( {P,\Pi } \right) &=& \frac{{\left| {\left( {Q - P} \right) \cdot N} \right|}}{{\left\| N \right\|}}  \cr 
  &=& \frac{{\left| {\left( {\left( {1,0,1} \right) - \left( {2,2,2} \right)} \right) \cdot \left( {2, - 1,2} \right)} \right|}}{{\sqrt {{2^2} + {{\left( { - 1} \right)}^2} + {2^2}} }}  \cr 
  &=&\frac{{\left| {\left( { - 1, - 2, - 1} \right) \cdot \left( {2, - 1,2} \right)} \right|}}{{\sqrt 9 }}  \cr 
  &=&\frac{{\left| { - 2 + 2 - 2} \right|}}{3} = \frac{2}{3} \cr} $$
Now I need to find $R$. But since $R$ is in $\Pi$, it must be that $R$ is the orthogonal projection of $P$ onto $\Pi$ so to speak, or, $R$ is the intersection of $\Pi$ with the line $$\mathscr L:\bar{x}  =\lambda\cdot(2,-1,2)+(2,2,2),\lambda\in \Bbb R$$
Thus, I get $$\eqalign{
  & x \in {\mathscr L} \Rightarrow x = (2\lambda  + 2,2 - \lambda ,2\lambda  + 2)  \cr 
  & x \in \Pi  \Rightarrow 2x - y + 2z = 4 \cr} $$ so
$$2\left( {2\lambda  + 2} \right) - \left( {2 - \lambda } \right) + 2\left( {2\lambda  + 2} \right) = 4$$
or $$\lambda  =  - \frac{2}{9}$$
Then $$R = \left( {\frac{{14}}{9},\frac{{20}}{9},\frac{{14}}{9}} \right)$$
So now I use as the normal vector of $\Pi'$ the cross product of $\bar{PR}$ and $\bar{PQ}$:
$$\eqalign{
   N &=& PQ \times PR =   \cr 
  &  =& \left( {Q - P} \right) \times \left( {R - P} \right)  \cr 
  &  =& \left( { - 1, - 2, - 1} \right) \times \left( { - \frac{4}{9},\frac{2}{9}, - \frac{4}{9}} \right)  \cr 
  &  =&  - \frac{2}{9}\left[ {\left( { - 1, - 2, - 1} \right) \times \left( {2, - 1,2} \right)} \right]  \cr 
  &  =&  - \frac{2}{9}\left( { - 4 - 1,2 - 2,1 + 4} \right) = \left( {\frac{{10}}{9},0,-\frac{{10}}{9}} \right) \cr} $$
So $\Pi'$ ends up being $$\Pi':{\frac{{10}}{9}x - \frac{{10}}{9}z = 0}$$ or
$$\boxed{ \Pi ':x = z} $$
Is this OK? Can it be checked graphically? Is there another way to solve this?

As you said $PR \perp \Pi$, and $R \in \Pi \cap \Pi'$.
Since $PR \subset \Pi'$ and $PR \perp \Pi$ you get $\Pi \perp \Pi'$ and hence the normal vector $(2,-1,2)$ of $\Pi$ is parallel to $\Pi'$.
But then the point $S$ so that $\vec{PS}=(2,-1,2)$ is in $\Pi'$.
You can easily find $S$, and then you can find the plane through $P,Q, S$ by standard methods. 
