= Solution
Let $a=B_{12}$, $b=B_{13}$, $c=B_{21}$, $d=B_{23}$, $e=B_{31}$, and $f=B_{32}$. Zero row sums put $B$ in the form used in the <spectrum of a three-dimensional matrix with zero row sums>. In particular, $B(1,1,1)^T=0$. Define
$$
S=a+b+c+d+e+f,
$$
$$
T=ad+bc+bd+ae+af+bf+ce+cf+de.
$$
Expanding the <characteristic polynomial> gives $\det(\lambda I-B)=\lambda(\lambda^2+S\lambda+T)$, so
$$
\boxed{\lambda_0=0,\qquad\lambda_\pm=\frac{-S\pm\sqrt{S^2-4T}}2.}
$$
For the physical coefficients, $L_jB_{jk}=L_kB_{kj}$ with $L_j=M_j\sqrt{GM_\star a_j}$. Thus $W=\operatorname{diag}(L_j)$ gives <positive diagonal symmetrization of a matrix>, and
$$
\mathbf z^\dagger WB\mathbf z=-\sum_{j<k}L_jB_{jk}|z_j-z_k|^2\le0.
$$
Consequently the <eigenvalues> are real and nonpositive, and $B$ has a complete <eigenvector> basis. This justifies the mode expansion in part (i).
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