= Solution
Put $E_1=E_2=E$. Within the degenerate eigenspace spanned by $|1\rangle,|2\rangle$, the perturbation is
$$
V_{\rm deg}
=\varepsilon V_0
\begin{pmatrix}0&1\\1&0\end{pmatrix}.
$$
<Degenerate perturbation theory> diagonalizes this matrix. Its normalized eigenstates $(|1\rangle\pm|2\rangle)/\sqrt2$ have first-order shifts $\pm\varepsilon V_0$. The coupling to $|3\rangle$ has order $\varepsilon^2$, so it cannot affect an energy through order $\varepsilon$. Therefore
$$
\boxed{
E_+=E+\varepsilon V_0,\qquad
E_-=E-\varepsilon V_0,\qquad
E_3'=E_3
\quad\text{through order }\varepsilon.}
$$
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