= Solution
Keep the inner-pair <Laplace coefficient> unexpanded because $\alpha=a_1/a_2$ is close to one. Put $\ell=L_1/L_2$ and
$$
\epsilon_B=\frac{M_3}{M_2}\frac{3\beta^3}{\alpha^2b_{3/2}^{(1)}(\alpha)},\qquad \beta=\frac{a_1}{a_3}\ll1.
$$
Using the small-argument <Laplace coefficient> for the distant planet and $n_2/n_1=\alpha^{3/2}$ gives
$$
\boxed{\begin{aligned}
B_{21}&=\ell B_{12},\\
B_{13}&\simeq\epsilon_B B_{12},\\
B_{23}&\simeq\alpha^{-3/2}\epsilon_B B_{12},\\
B_{31}&\simeq\frac{L_1}{L_3}\epsilon_B B_{12},\\
B_{32}&\simeq\frac{L_2}{L_3}\alpha^{-3/2}\epsilon_B B_{12}.
\end{aligned}}
$$
The reverse-coupling expressions follow from the <angular momentum> identity $L_jB_{jk}=L_kB_{kj}$. If desired, $L_1/L_2=(M_1/M_2)\sqrt\alpha$, $L_1/L_3=(M_1/M_3)\sqrt\beta$, and $L_2/L_3=(M_2/M_3)\sqrt{\beta/\alpha}$ express all ratios in the supplied quantities. The diagonal entries are $B_{11}=-(B_{12}+B_{13})$, $B_{22}=-(B_{21}+B_{23})$, and $B_{33}=-(B_{31}+B_{32})$.
For comparable masses and closely spaced inner orbits, $\alpha\simeq1$, $\ell\sim1$, and $L_1/L_3,L_2/L_3=O(\sqrt\beta)$. Thus
$$
\boxed{B_{12}\sim B_{21}\gg B_{13}\sim B_{23}\gg B_{31}\sim B_{32}.}
$$
The distant couplings $B_{13},B_{23}$ scale as $\beta^3$ relative to a fixed inner architecture, and the reverse couplings have an additional $\sqrt\beta$ suppression.
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