= Solution
Hermiticity supplies the reverse <matrix elements>, also equal to the real number $t$. In an <orthonormal basis> for the degenerate <eigenspace>, the perturbation is
$$
H_1=t\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}
=t(\mathbf J_3-I),
$$
where $\mathbf J_3$ is the <all-ones matrix>. Its symmetric <vector> $(1,1,1)/\sqrt3$ has <eigenvalue> $2t$. Every <vector> whose components sum to zero has <eigenvalue> $-t$; that subspace has dimension two. For example, $(1,-1,0)/\sqrt2$ and $(1,1,-2)/\sqrt6$ form an <orthonormal basis> for it.
The first-order energies are therefore
$$
\boxed{E_0+2t\quad\text{once},\qquad E_0-t\quad\text{twice}.}
$$
The <trace> of the shifts is $2t-t-t=0$, agreeing with the zero diagonal. The ordering depends on the sign of $t$; the formula does not assume that the symmetric state is the lowest. Diagonalizing the perturbation inside the degenerate space is essential: the original diagonal entries alone would incorrectly predict no splitting.
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