You can easily check for A considering the product by the basis vector of the plane, since $\forall v$ in the plane must be:
$$Av=v$$
Whereas for the normal vector:
$$An=0$$
Note that with respect to the basis $\mathcal{B}:{c_1,c_2,n}$ the projection matrix is simply:
$$P_{\mathcal{B}}=\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\end{bmatrix}$$
If you need the projection matrix with respect to another basis you simply have to apply a change of basis to obtain the new matrix.
For example with respect to the canonical basis, lets consider the matrix M which have vectors of the basis $\mathcal{B}:{c_1,c_2,n}$ as colums:
$$M=\begin{bmatrix}
-1 & 0 & 1\\
0 & -1 & 1\\
1 & 1 & 1\end{bmatrix}$$
If w is a vector in the basis $\mathcal{B}$ its expression in the canonical basis is $v$ give by:
$$v=Mw\implies w=M^{-1}v$$
Thus if the projection $w_p$ of w in the basis $\mathcal{B}$ is given by:
$$w_p=P_{\mathcal{B}}w$$
The projection in the canonical basis is given by:
$$M^{-1}v_p=P_{\mathcal{B}}M^{-1}v\implies v_p=MP_{\mathcal{B}}M^{-1}v $$
Thus the matrix:
$$A=MP_{\mathcal{B}}M^{-1}=$$
$$=\begin{bmatrix}
-1 & 0 & 1\\
0 & -1 & 1\\
1 & 1 & 1\end{bmatrix}\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\end{bmatrix}\begin{bmatrix}
-1 & \frac13 & \frac13\\
\frac13 & -1 & \frac13\\
\frac13 & \frac13 & \frac13\end{bmatrix}=\begin{bmatrix}
2/3 & -1/3 & -1/3\\
-1/3 & 2/3 & -1/3\\
-1/3 & -1/3 & 2/3\end{bmatrix}$$
represent the projection matrix in the plane with respect to the canonical basis.
Suppose now we want find the projection matrix from the base $\mathcal{B}$ to the canonical $\mathcal{C}$.
Let's consider the projection $w_p$ of w in the basis $\mathcal{B}$ is given by:
$$w_p=P_{\mathcal{B}}w$$
thus:
$$M^{-1}v_p=P_{\mathcal{B}}w\implies v_p=MP_{\mathcal{B}}w$$
Thus the matrix:
$$C=MP_{\mathcal{B}}=$$
$$=\begin{bmatrix}
-1 & 0 & 1\\
0 & -1 & 1\\
1 & 1 & 1\end{bmatrix}\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\end{bmatrix}=\begin{bmatrix}
-1 & 0 & 0\\
0 & -1 & 0\\
1 & 1 & 0\end{bmatrix}$$
represent the projection matrix from the base $\mathcal{B}$ to the canonical $\mathcal{C}$.
Best Answer
Assuming you mean the orthogonal projection onto the plane $W$ given by the equation $x-y-z$, it is equal to the identity minus the orthogonal projection onto $W^\perp$, which is sightly easier to compute. Now $W^\perp$ is the span of the normal vector $v=(1,-1,-1)$, and the orthogonal projection onto which is $x\mapsto \frac{(v\mid x)}{(v\mid v)}v$, and whose matrix is $$ \frac13\begin{pmatrix}1\\-1\\-1\end{pmatrix} \begin{pmatrix}1&-1&-1\end{pmatrix} =\frac13\begin{pmatrix}1&-1&-1\\-1&1&1\\-1&1&1\end{pmatrix}. $$ Subtracting this from the identity gives $$ \begin{pmatrix}2/3&1/3&1/3\\1/3&2/3&-1/3\\1/3&-1/3&2/3\end{pmatrix}. $$