= Solution
Projecting $H|\Psi\rangle=E|\Psi\rangle$ onto $\langle\psi_n|$ and using orthonormality gives
$$
E_0c_n+\sum_{m=0}^2\langle\psi_n|V|\psi_m\rangle c_m=Ec_n.
$$
Thus, with $d=E_0+\alpha$,
$$
\begin{pmatrix}
d&-A&-A\\
-A&d&-A\\
-A&-A&d
\end{pmatrix}
\begin{pmatrix}c_0\\c_1\\c_2\end{pmatrix}
=E\begin{pmatrix}c_0\\c_1\\c_2\end{pmatrix}.
$$
The symmetric vector $(1,1,1)$ has eigenvalue
$$
E_s=d-2A=E_0+\alpha-2A.
$$
Let $\omega=e^{2\pi i/3}$, so $1+\omega+\omega^2=0$. For either vector
$$
(1,\omega,\omega^2),
\qquad
(1,\omega^2,\omega),
$$
the other two components in each row sum to minus the component in that row. Both therefore have eigenvalue
$$
E_d=d+A=E_0+\alpha+A.
$$
The spectrum is consequently
$$
\boxed{E_0+\alpha-2A,\quad E_0+\alpha+A,\quad E_0+\alpha+A},
$$
with normalized eigenvectors obtained by multiplying the three displayed vectors by $1/\sqrt3$.
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