i think it is easier to find the projection on to the line $u = (1, -2, 3)^\top$ that is orthogonal to the plane and then subtract from the identity to get the projection onto the plane.
the projection matrix onto the line $u$ is $$uu^\top/(u^\top u) = \frac1{14}\pmatrix{1&-2&3\\-2&4&-6\\3&-6&9}.$$ therefore the projection matrix on to the plane is $$I - uu^\top/(u^\top u) = \frac1{14}\pmatrix{13&2&-3\\2&10&6\\-3&6&5}.$$
You are right: $(1,1,1)$ is orthogonal to $V$. Therefore, $A.(1,1,1)=(0,0,0)$. Now, consider the vectors $(1,-1,0)$ and $(1,0,-1)$. Since they both belong to $V$, you must have $A.(1,-1,0)=(1,-1,0)$ and $A.(1,0,-1)=(1,0,-1)$.
Now, since$$(1,0,0)=\frac13(1,1,1)+\frac13(1,-1,0)+\frac13(1,0,-1),$$you must have$$A.(1,0,0)=\frac13(1,-1,0)+\frac13(1,0,-1)=\left(\frac23,-\frac13,-\frac13\right).$$So, the entries of the first column of the matrix of $\operatorname{proj}_V$ with respect to the standard basis will be $\frac23$, $-\frac13$ and $-\frac13$. Can you take it from here?
Best Answer
The first two vectors $e_x$ and $e_y$ are invariant under the projection, and the last one is mapped to 0.
Hence the columns of the matrix are, in order;
$$ \begin{pmatrix} 1\\ 0 \\0 \end{pmatrix} ; \begin{pmatrix} 0\\ 1 \\0 \end{pmatrix} ; \begin{pmatrix} 0\\ 0 \\0 \end{pmatrix} $$