 Solution

The states $|2⟩$ and $|3⟩$ are degenerate and couple to $|1⟩$ with amplitudes $\epsilon V_0$ and ${\epsilon }^2{V}_0$. The bright and dark state reduction of a star-coupled Hamiltonian gives one exact dark eigenstate proportional to $-\epsilon |2⟩+|3⟩$, with energy $E$. The orthogonal bright state couples to $|1⟩$ with magnitudeDiagonalizing the remaining two-dimensional Hermitian matrix gives the exact spectrumWriting $\Delta =E_1-E\ne 0$ and expanding the branches according to the eigenvalue to which they converge,These are exactly the three perturbative energies from part b through order ${\epsilon }^2$.