= Solution
Put $d=\ell/(2\sqrt3)$ and $a=e^{2\pi i/3}$. Consistent with the actual PDF vertices, parametrize the three oriented sides by
$$
z^{(j)}(s)=\beta_j(d+is),\qquad (\beta_1,\beta_2,\beta_3)=(1,\bar a,a),\qquad-\ell/2\leq s\leq\ell/2.
$$
They run counterclockwise, with unit tangent $i\beta_j$ and outward normal $\beta_j$. Consequently the <normal derivative> and tangential derivative satisfy
$$
q_s=i\beta_jq_z-i\bar\beta_jq_{\bar z},\qquad q_N=\beta_jq_z+\bar\beta_jq_{\bar z},\qquad i\beta_jq_z=\frac{q_s+iq_N}{2}.
$$
Let $\kappa=k\beta_j$. On this side the exponential factor becomes
$$
\chi=E(-i\kappa)e^{(\kappa+\lambda/\kappa)s},\qquad E(k)=e^{(k+\lambda/k)d}.
$$
Pulling back the <closed differential one-form> therefore gives
$$
W|_j=E(-i\kappa)e^{(\kappa+\lambda/\kappa)s}\left[\frac12q_s^{(j)}+\frac\lambda\kappa q^{(j)}+\frac i2q_N^{(j)}\right]ds.
$$
Define $\Phi_j(\kappa)$ and $\Psi_j(\kappa)$ by integrating the Dirichlet terms and normal term in this expression. The <Generalized Stokes theorem> gives $\int_{\partial D}W=0$, so
$$
\boxed{E(-ik)\Psi_1(k)+E(-i\bar ak)\Psi_2(\bar ak)+E(-iak)\Psi_3(ak)=2i\bigl[E(-ik)\Phi_1(k)+E(-i\bar ak)\Phi_2(\bar ak)+E(-iak)\Phi_3(ak)\bigr].}
$$
The factor $2i$ follows by solving $\sum E\Phi+(i/2)\sum E\Psi=0$. The bars on the side-2 rotation are essential; replacing them with $a$ would use the wrong side orientation. The converted TeX also reverses the sign in the upper vertex's exponential, whereas the PDF puts that vertex above the real axis; the parametrization here follows the PDF.
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