Question: We got

$P\_1 : (1,1,0)$
$P\_2 : (2,2,2)$
$P\_3 : (2,3,1)$
$P\_4 : (3,2,1)$

And should calculate the angle between the normal vector to the triangle of $P\_1, P\_2, P\_3$ and $\boldsymbol{P\_1P\_4}$.

I've started with $P\_1$ in the plane equation to get the normal:
$\pi = 1a+1b+2c=0 \Rightarrow (1,1,-1)$. I think my error is here.

I then use the formula for the angle between two vectors:
$ cos(\boldsymbol{u},\boldsymbol{v}) = \frac{\boldsymbol{u} \cdot \boldsymbol{v}}{|\boldsymbol{u}||\boldsymbol{v}|}$ and that gives me $\frac{2-2}{\sqrt{3}\sqrt{6}}$ Which is clearly wrong. The correct answer should be $\frac{4}{\sqrt{66}}$.

Answer: 
The plane normal is obtained by:
$$
\boldsymbol{n} = \left(P\_2 - P\_1\right) \times \left(P\_3 - P\_1\right) \\
= (1,1,2)\times(1,2,1) \\
= (-3,1,1) \\
\Rightarrow \hat{\boldsymbol{n}} = \frac{(-3,1,1)}{\sqrt{11}}\\
P\_1P\_4 \equiv \boldsymbol{v} = P\_4 - P\_1 = (2, 1, 1) \Rightarrow \hat{\boldsymbol{v}} = \frac{(2,1,1)}{\sqrt{6}}\\
\therefore cos\theta = \hat{\boldsymbol{n}} \cdot \hat{\boldsymbol{v}} = -\frac{4}{\sqrt{66}}
$$
I'm assuming your $P\_1P\_4$ points in the opposite direction from what I've assumed here, that would account for the minus sign in the above answer.
This is pretty straightforward . You dont even need to find the equation of the plane to do this calculation . $P\_1P\_2 \times P\_1P\_3$ will give you the direction of the normal vector to the plane that contains the triangle $P\_1P\_2P\_3$ . Then you can use the formula 

$$ cos(\theta) =\frac{u\cdot v}{|u||v|} $$