= Solution
Set
$$
q=\frac{im\lambda}{\hbar^2k},\qquad r=e^{ika},\qquad
\boldsymbol\Psi=\begin{pmatrix}\psi(-a)\\\psi(0)\\\psi(a)\end{pmatrix}.
$$
The <integral> equation becomes
$$
\psi(x)=e^{ikx}+q\left\{e^{ik|x+a|}\psi(-a)+e^{ik|x|}\psi(0)+e^{ik|x-a|}\psi(a)\right\}.
$$
Evaluating it at the three delta <functions> gives
$$
\begin{aligned}
\psi(-a)&=r^{-1}+q\{\psi(-a)+r\psi(0)+r^2\psi(a)\},\\
\psi(0)&=1+q\{r\psi(-a)+\psi(0)+r\psi(a)\},\\
\psi(a)&=r+q\{r^2\psi(-a)+r\psi(0)+\psi(a)\}.
\end{aligned}
$$
Thus
$$
\left[I-q\begin{pmatrix}1&r&r^2\\r&1&r\\r^2&r&1\end{pmatrix}\right]
\boldsymbol\Psi
=\begin{pmatrix}r^{-1}\\1\\r\end{pmatrix}.
$$
For $x>a$, all three Green-function terms are proportional to $e^{ikx}$. Hence
$$
S_{++}(k)=1+q\left\{r\psi(-a)+\psi(0)+r^{-1}\psi(a)\right\}.
$$
Now define $\gamma=ik\hbar^2/(\lambda m)$, so that $q=-1/\gamma$. If $M$ denotes the <matrix> multiplying $\boldsymbol\Psi$, direct evaluation gives
$$
\det M=\frac{r^4}{\gamma^3}\left[1-\gamma-2r^{-2}(1+\gamma)+r^{-4}(1+\gamma)^3\right].
$$
The displayed algebraic equation is therefore exactly the condition that the finite-dimensional scattering system become singular. Its solutions are poles of the analytically continued <scattering amplitude>.
To see the imaginary-axis poles directly, write $k=i\kappa$ with $\kappa>0$. When $a\to0$, the equation reduces to
$$
\gamma^2(\gamma+3)=0,
$$
and its nonzero root is $\gamma=-3$, or
$$
\kappa=\frac{3m\lambda}{\hbar^2}.
$$
This is the <bound state> of the coincident potential $-3\lambda\delta(x)$. When $a\to\infty$, the three centres decouple and the roots approach
$$
\gamma=-1,\qquad \kappa=\frac{m\lambda}{\hbar^2},
$$
with exponentially small splitting. A pole at $k=i\kappa$ has energy $E=-\hbar^2\kappa^2/(2m)$ and an exponentially decaying <wavefunction>, so these upper imaginary-axis singularities represent <bound states>.
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