= Solution
Work with real positive pulse envelopes and define $\Omega=\sqrt{\Omega_{12}^2+\Omega_{23}^2}>0$. Then $\cos\theta=\Omega_{12}/\Omega$ and $\sin\theta=\Omega_{23}/\Omega$. In the ordered basis $|1\rangle,|2\rangle,|3\rangle$, the candidate <dark state of a driven Hamiltonian> has coordinates
$$
|D\rangle=\frac1\Omega\begin{pmatrix}-\Omega_{23}\\0\\\Omega_{12}\end{pmatrix}.
$$
Multiplication by the given <rotating-wave approximation> matrix gives
$$
H^{\rm RWA}|D\rangle=\frac1\Omega\begin{pmatrix}0\\-\Omega_{12}\Omega_{23}+\Omega_{23}\Omega_{12}\\0\end{pmatrix}=0.
$$
Thus
$$
\boxed{|\Psi_0\rangle=\cos\theta|3\rangle-\sin\theta|1\rangle,\qquad\lambda_0=0.}
$$
Its absence of an excited-state component is the destructive interference that makes it dark. The orthogonal bright ground-state combination is $|B\rangle=\cos\theta|1\rangle+\sin\theta|3\rangle$. Since $H|B\rangle=\Omega|2\rangle$ and $H|2\rangle=\Omega|B\rangle$, the two remaining eigenstates are $(|B\rangle\pm|2\rangle)/\sqrt2$ with <eigenvalues> $\pm\Omega$.
For signed envelopes, use a continuous two-argument angle with $\cos\theta=\Omega_{12}/\Omega$ and $\sin\theta=\Omega_{23}/\Omega$, rather than an ambiguous arctangent branch. When both fields vanish, every state has zero <eigenvalue> and this formula does not select a unique dark vector.
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