Fact: $P$ is a projection matrix iff $P^2 = P$.
So, we need to show that $P^2 = P \implies (I-P)^2 = I-P$.
Do you see how to do this?
EDIT: As mentioned by Vedran Šego in the comments below, the above only shows that $P$ is a projection matrix, not necessarily an orthogonal projection matrix. To show that $P$ is an orthogonal projection matrix, we also need to show that $P$ is symmetric $\implies$ $I-P$ is symmetric.
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
The simplest way to compute this matrix is to consider how you would normally compute this projection. Let $v$ denote the normal vector $v = (0,-1,-2,-1)$. For a vector $x$, the projection onto the orthogonal complement $W^\perp$ of the hyperplane is given by $$ \operatorname{proj}_{v}(x) = \frac{v^Tx}{v^Tv}\cdot v. $$ So, the projection of $x$ onto $W$ is given by $x - \operatorname{proj}_{v}(x)$. Our goal is to produce a matrix $P$ for which $$ Px = x - \operatorname{proj}_{v}(x) = x - \frac{v^Tx}{v^Tv}\cdot v. $$ A nice way to find this matrix $P$ is to factor the right-hand side into the form $Mx$ for some matrix $M$. In particular, note that $$ x - \frac{v^Tx}{v^Tv}\cdot v = x - \frac 1{v^Tv}v\cdot v^Tx = x - \frac 1{v^Tv}vv^T x = \left(I - \frac{vv^T}{v^Tv} \right)x, $$ where $I$ dentoes the identity matrix. So, our matrix $P$ should be given by $P = I - \frac{vv^T}{v^Tv}.$ If you compute this correctly, you should end up with the answer $$ P = \pmatrix{1 & 0 & 0 & 0\\ 0 & \frac{5}{6} & -\frac{1}{3} & -\frac{1}{6}\\ 0 & -\frac{1}{3} & \frac{1}{3} & -\frac{1}{3}\\ 0 & -\frac{1}{6} & -\frac{1}{3} & \frac{5}{6}}. $$