(This discussion applies to finite dimensional spaces. Recall that to define a linear operator, it is sufficient to define its behaviour on a basis.)
Your approach is correct. However, there is some ambiguity in defining $P$, you have defined one projection, but there are others.
Suppose you have a projection $P$ onto a subspace $A$ with basis $a_1,.., a_k$. Then $P$ is uniquely defined on $A$ (since $Pa_i = a_i$). Suppose $b_{k+1},.., b_n$ together with $a_1,.., a_k$ form a basis for the whole space. Then $P$ can be arbitrarily defined on $b_i$ as long as $P b_i \in A$. So, the projection is not unique.
Back to the problem on hand:
Let $v_1,v_2,v_3$ be the vectors you have in $\mathscr{C}$ which form a basis for $\mathbb{R}^3$ (and $v_1, v_2$ form a basis for $A$). Then you must have
$P v_1 = v_1$, $P v_2 = v_2$, but the only requirement for $P v_3$ is that it lie in $A$. So $P v_3 = \alpha_1 v_1 + \alpha_2 v_2$, where $\alpha_i$ are arbitrary (but fixed, of course).
Let $V = \begin{bmatrix} v_1&v_2&v_3 \end{bmatrix}$. Then we have $PV = \begin{bmatrix} v_1&v_2& \alpha_1 v_1 + \alpha_2 v_2\end{bmatrix} = W$.
You were asked to compute $P e_i$, which is tantamount to computing $P = W V^{-1}$. Noting that $W = \begin{bmatrix} v_1 & v_2 & 0 \end{bmatrix} + \alpha_1 \begin{bmatrix} 0 & 0 & v_1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 0 & v_2 \end{bmatrix}$, we see that P can be expressed as
$$ P = \begin{bmatrix} v_1 & v_2 & 0 \end{bmatrix}V^{-1} + \alpha_1 \begin{bmatrix} 0 & 0 & v_1 \end{bmatrix} V^{-1} + \alpha_2 \begin{bmatrix} 0 & 0 & v_2 \end{bmatrix} V^{-1}$$
Grinding through through the calculation gives:
$$P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 2 & 0 \end{bmatrix} + \alpha_1 \begin{bmatrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ -1 & 2 & -1 \end{bmatrix} + \alpha_2 \begin{bmatrix} 0 & 0 & 0 \\ 1 & -2 & 1 \\ 2 & -4 & 2 \end{bmatrix}$$
For 1: $v_2$ and $v_3$ are constructed to be in the plane under question. Thus, the projection leaves these vectors as they are, $T(v_2)=v_2$ and $T(v_3)=v_3$.
(Expanded: These vectors are already in the plane on which $T$ projects. The projection onto a plane takes vectors outside the plane and throws them onto the plane. If such a vector is already on a plane, then the projection does nothing.)
Moreover, $v_1$ is the direction normal to the plane, hence $T(v_1)=0$.
For 2: With similar considerations, one finds $T(v_1)=-v_1$, $T(v_2)=v_2$ and $T(v_3)=v_3$.
(Expanded: Reflection changes the direction of incoming rays. If the ray is perpendicular to the plane (or the mirror) it gets thrown back and its direction changes: $T(v_1)=-v_1$.)
Best Answer
Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $\mathbf e_1$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $\mathbf e_1$ is equal to $\mathbf e_1-\mathbf n = (-1,-2,-1)^T$. Unfortunately, this doesn’t even lie on the plane: $2(-1)+2(-2)+1(-1)=-7$.
The problem is that you’ve set the rejection of $\mathbf e_1$ from the plane to be equal to $\mathbf n$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $P\mathbf e_1$ of $\mathbf e_1$ onto the plane is $\mathbf e_1-k\mathbf n$ for some as-yet-undetermined scalar $k$. However, $k\mathbf n$ here is simply the orthogonal projection of $\mathbf e_1$ onto $\mathbf n$, which I suspect that you know how to compute.