= Solution
At the no-convection <equilibrium> $(a,b,c)=(0,0,0)$, one <eigenvalue> is $-\zeta$ and the other two come from
$$
J_{ab}=\begin{pmatrix}\sigma(r-1)&-\sigma\zeta q\\1&-\zeta\end{pmatrix}.
$$
Their <trace> and <determinant> are
$$
T=\sigma(r-1)-\zeta,\qquad D=\sigma\zeta(1+q-r).
$$
A zero <eigenvalue> occurs at $r=1+q$. If $q<\zeta/\sigma$, the other two <eigenvalues> are negative, and the stationary <bifurcation> is the pitchfork analyzed in part (c). For $q>\zeta/\sigma$ the transverse <eigenvalue> is positive instead.
An imaginary pair occurs at $r=1+\zeta/\sigma$, provided $q>\zeta/\sigma$, because
$$
\omega_H^2=D=\sigma\zeta q-\zeta^2>0.
$$
The <trace> crosses zero with <derivative> $\sigma>0$. The cubic coupling is nondegenerate here: at criticality, putting $K=\sigma\zeta q$ reduces the oscillatory variables to
$$
a''+\omega_H^2a=Kac,\qquad
c'=-\zeta c+\frac3K(\zeta a^2-aa').
$$
For $a=R\cos(\omega_H\tau)$, the forced second harmonic of $c$ has sine coefficient $9R^2\zeta\omega_H/[2K(\zeta^2+4\omega_H^2)]$. Averaging the <derivative> of $(a'^2+\omega_H^2a^2)/2$ consequently gives $-9\zeta\omega_H^2R^4/[8(\zeta^2+4\omega_H^2)]<0$. Thus the <Hopf bifurcation> is nondegenerate and supercritical.
The two thresholds meet at
$$
\boxed{q=\frac\zeta\sigma,\qquad r=1+\frac\zeta\sigma.}
$$
Here the planar block is nonzero but has <trace> and <determinant> zero, while the third <eigenvalue> remains negative. It is a reflection-symmetric double-zero, codimension-two point; the ordinary simple-eigenvalue reduction of part (c) is not valid there.
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