Matrix associated to a projection onto the plane $π$ with equation $x−y + 2z = 0$ Let $T \colon  \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear transformation which projects vectors onto the plane $π$ with equation $x−y + 2z = 0$. Find a matrix $A$ such that $T = T_A$  
My workings so far:  
$$ A = 
\left(\begin{bmatrix}
T(e_1)&T(e_2)&T(e_3)
\end{bmatrix}\right).  $$
First I need to find $f_1,f_2$ that span the plane $\pi$ and they are orthogonal. So I chose 
$$ f_1 = 
\left(\begin{bmatrix}
1
\\1  
\\0
\end{bmatrix}\right),  f_2 = 
\left(\begin{bmatrix}
-1
\\1  
\\1
\end{bmatrix}\right). $$
Next I need to use this formula but I am not quite sure how to
$$T(u)= uf_1/[f_1f_1] f_1 + uf_2/[f_2f_2] f_2$$
Looking for some help with carrying out the rest of the problem. Sorry I did not know the proper formatting for the $T(u)$ formula

You have
$$ T \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{\left< \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \right>}{\left \| \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \right \|^2 }\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}  +  \frac{\left< \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \right>}{\left \| \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \right \|^2 }\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} = \frac{x+y}{2} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \frac{-x + y + z}{3} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \\
= x \begin{pmatrix} \frac{5}{6} \\ \frac{1}{6} \\ -\frac{1}{3} \end{pmatrix} + y \begin{pmatrix} \frac{1}{6} \\ \frac{5}{6} \\ \frac{1}{3} \end{pmatrix} + z \begin{pmatrix} -\frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \end{pmatrix} = \underbrace{\begin{pmatrix} \frac{5}{6} & \frac{1}{6} & -\frac{1}{3} \\ \frac{1}{6} & \frac{5}{6} & \frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{pmatrix}}_{A} \begin{pmatrix} x \\ y \\ z \end{pmatrix}.   $$
