The Matrix of a reflection (around abitrary plane) Let $\Upsilon :\mathbb{R}^3\rightarrow \mathbb{R}^3$ be a reflection across the plane: $\pi : -x + y + 2z = 0 $. Find the matrix of this linear transformation using the standard basis vectors and the matrix which is diagonal.
Now first of, If I have this plane then for $\Upsilon(x,y,z) = (-x,y,2z)$ I get this when passing any vector, so the matrix using standard basis vectors is:
$$Y=\begin{pmatrix}-1&0&0\\ 0&1&0\\ 0&0&-2\end{pmatrix}\;.$$
I feel like this is totally wrong and what If I had like $ax+by+cz=0$ an arbitrary plane?
 A: If $n$ is the plane's unit normal vector (here $(-1,1,2)/\sqrt6$), then the reflection of any vector $p$ across the plane is given by
$$p-2\langle n,p\rangle n.$$
Now plug in the vectors of the standard basis to get the column vectors of the desired matrix.
Facit: avoid coordinates.
Edit: if $n$ is not a unit vector we have
$$p-\frac{2\langle n,p\rangle}{\|n\|^2}n.$$
A: I’m not quite sure what you’re asking, but here’s a way to construct the matrix of the reflection via a diagonal matrix:  
A reflection of a vector across a plane (more generally, across an $(n-1)$-dimensional subspace of an $n$-dimensional space) reverses the component of the vector that’s orthogonal to the plane and leaves fixed its component in the plane. Another way of putting this is that the reflection is the identity on vectors in the plane and multiplication by $-1$ on vectors orthogonal to it. So, take a basis for the plane and extend it by adding a vector normal to it. Relative to this basis, the matrix of the reflection is simply $$Y=\pmatrix{1&0&0\\0&1&0\\0&0&-1}$$ from which you can get $\Upsilon$ via a change of basis: $BYB^{-1}$. Note that right-multiplication by $Y$ just changes the sign of the third column of a matrix.  
In your case, the equation of the plane immediately gives us the normal $(-1,1,2)^T$. A basis for the plane is easily generated from this vector: $\{(1,1,0)^T,(2,0,1)^T\}$. Putting this all together gives $$\Upsilon=\pmatrix{1&2&1\\1&0&-1\\0&1&-2}\pmatrix{1&2&-1\\1&0&1\\0&1&2}^{-1}=\pmatrix{\frac23&\frac13&\frac23\\\frac13&\frac23&-\frac23\\\frac23&-\frac23&-\frac13}.$$
The construction in Michael Hoppe’s answer is easier to calculate, though, especially in higher-dimensional spaces.
