= Solution
The two <invariant subspaces> are
$$
U_0=\mathbb C(1,1,1)^T,\qquad
U_2=\{(z_1,z_2,z_3)^T:z_1+z_2+z_3=0\}.
$$
Coordinate <permutations> preserve both equal coordinates and the coordinate sum. Their <dimensions> add to three and their intersection is zero, so $\mathbb C^3=U_0\oplus U_2$. On $U_0$, the <basis> $u_0=(1,1,1)^T$ gives $R_{U_0}(\tau)=(1)$ for every <transposition>.
On $U_2$ choose the <basis> $u=e_1-e_3$, $v=e_2-e_3$. Directly permuting their entries gives
$$
\begin{array}{c|cc}
\tau&R(\tau)u&R(\tau)v\\\hline
(12)&v&u\\
(13)&-u&v-u\\
(23)&u-v&-v
\end{array}
$$
Thus, taking images as columns, the <matrices> of the <matrix representation> are
$$
\boxed{D(12)=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad
D(13)=\begin{pmatrix}-1&-1\\0&1\end{pmatrix},\quad
D(23)=\begin{pmatrix}1&0\\-1&-1\end{pmatrix}.}
$$
Each has <trace> zero and squares to $I$. Their three-cycle products have <trace> $-1$, so this is exactly the two-dimensional <irreducible representation> of <2/a/solution>. These are all nonzero <invariant subspaces> affording individual <irreducible representations>, since each occurs once in <2/b/solution>.
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