= Solution
Fix counterclockwise arclength on each side and take the outward <normal derivative>. Write
$$
h=\frac l{2\sqrt3},\qquad z^{(j)}(s)=c_j(h+is),\qquad
(c_1,c_2,c_3)=(1,\bar a,a),\quad a=e^{2\pi i/3}.
$$
The unit tangent is $ic_j$ and the outward unit normal is $c_j$. On side $j$ this gives the chain-rule identity
$$
i c_jq_z=\frac12q_s^{(j)}+\frac i2q_N^{(j)}.
$$
Put $w=c_jk$. The exponential in the <closed differential one-form> pulls back to
$$
W(z^{(j)}(s),k)=E(-iw)e^{(w+\lambda/w)s},\qquad
\boxed{E(k)=\exp\left[\frac l{2\sqrt3}\left(k+\frac\lambda k\right)\right].}
$$
Also $-\lambda q\,d\bar z/(ik)=(\lambda/w)q^{(j)}ds$. Consequently the contribution of side $j$ is $\rho_j(c_jk)$, where the required local side spectra are
$$
\boxed{\begin{aligned}
\Psi_j(k)&=\int_{-l/2}^{l/2}e^{(k+\lambda/k)s}q_N^{(j)}(s)ds,\\
\Phi_j(k)&=\int_{-l/2}^{l/2}e^{(k+\lambda/k)s}
\left[\frac12q_s^{(j)}(s)+\frac\lambda kq^{(j)}(s)\right]ds,\\
\rho_j(k)&=E(-ik)\left[\frac i2\Psi_j(k)+\Phi_j(k)\right].
\end{aligned}}
$$
These are spectral <integral transforms> in the local coordinate of each side. Keeping the tangential-derivative term is important: premature <integration by parts> in a single side integral would leave corner values, which cannot simply be discarded. At the special samples used below, those endpoint terms do vanish for the rotationally invariant solution.
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