Orthogonal projection question 
Consider the (orthogonal) projection $T: \mathbb{R}^3 \to \mathbb{R}^3$ onto the plane $x - y + z = 0$.
  
  
*
  
*(a) Find the standard matrix $[T]_S$ for $T$.
  
*(b) Find a new basis $B$ so that $[T]_B$ is rather simple.

How do I do this question?
I created an orthogonal basis and projected $(x,y,z)$.
My result was 
$[T]_s=\begin{bmatrix}11/18 & 7/18 & -2/9\\
7/18 & 11/18 & 2/9\\
-2/9 & 2/9 & 4/9
\end{bmatrix}$
and what about part b? Do I make the new basis equal to the orthogonal one I found when solving part a? and is $[T]_B$ a $3 \times 3$ unit matrix?
 A: Your matrix is wrong. An easy sanity check is that a matrix representing an orthogonal projection should be symmetric with trace equal to the dimension of the space you are projecting onto (in your case, two).
A unit normal vector to your plane is $v = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and so the projections of the standard basis vectors are given by
$$ T(e_1) = e_1 - \left<e_1, v \right>v = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} - \frac{1}{\sqrt{3}} \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} \\ \frac{1}{3} \\ -\frac{1}{3} \end{pmatrix}, \\
T(e_2) = e_2 - \left<e_2, v \right>v = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + \frac{1}{3} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ \frac{2}{3} \\ \frac{1}{3} \end{pmatrix}, \\
T(e_3) = e_3 - \left<e_3, v \right>v = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} - \frac{1}{\sqrt{3}} \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} =  \begin{pmatrix} -\frac{1}{3} \\ \frac{1}{3} \\ \frac{2}{3} \end{pmatrix} $$
and so
$$ [T]_S = \begin{pmatrix} [Te_1]_S & [Te_2]_S & [Te_3]_S \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & \frac{1}{3} & -\frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} & \frac{2}{3} \end{pmatrix}. $$
If you pick an orthonormal basis $(v_1,v_2)$ for the plane $x - y + z = 0$ and take $v_3 = v$, then with respect to the new basis, the matrix will become
$$ [T]_{(v_1,v_2,v_3)} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$
