= Solution
Now put $E_2=E_3=E$ and $E_1\ne E$. All diagonal matrix elements of $V$ vanish. The leading coupling is $\langle1|V|2\rangle=\varepsilon V_0$, whereas $\langle1|V|3\rangle=\varepsilon^2V_0$. Applying the second-order formula gives
$$
\boxed{
\begin{aligned}
E_1'&=E_1+\frac{\varepsilon^2V_0^2}{E_1-E}+O(\varepsilon^4),\\
E_2'&=E+\frac{\varepsilon^2V_0^2}{E-E_1}+O(\varepsilon^4),\\
E_3'&=E+O(\varepsilon^4).
\end{aligned}}
$$
The degeneracy causes no singular denominator through this order because $\langle2|V|3\rangle=0$. Equivalently, the effective perturbation on the degenerate subspace begins as
$$
\frac{V_0^2}{E-E_1}
\begin{pmatrix}
\varepsilon^2&\varepsilon^3\\
\varepsilon^3&\varepsilon^4
\end{pmatrix}.
$$
Its off-diagonal mixing is only order $\varepsilon^3$, so \b[degenerate perturbation theory is unnecessary for energies through order $\varepsilon^2$.]
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