= Solution
For low <temperature>, factor out the largest <Boltzmann factor> from the dominant <eigenvalue>:
$$
\lambda_+=3e^{2K}\left(1+e^{-2K}+\frac13e^{-4K}+\frac13e^{-6K}\right).
$$
Thus the <Helmholtz free energy> per site has the expansion
$$
f=-\frac{2J}{3}-\frac{k_BT}{3}\ln3-\frac{k_BT}{3}e^{-2K}+O\!\left(k_BT e^{-4K}\right).
$$
Differentiate at fixed $J$, remembering that $K=J/(k_BT)$. The <entropy> per site is
$$
s=\frac{k_B}{3}\ln3+\frac{k_B}{3}(1+2K)e^{-2K}+O\!\left(k_B(1+K)e^{-4K}\right),
\qquad \boxed{\lim_{T\to0^+}s(T)=\frac{k_B}{3}\ln3.}
$$
The nonzero <residual entropy> comes from an exponentially large <ground-state degeneracy>, generated by <geometrical frustration>. A triangle with antiferromagnetic <energy> $J[st+(s+t)\tau]$ has minimum <energy> $-J$: if $s=t$, the apex must have the opposite sign; if $s=-t$, either apex sign is permitted. The two all-equal configurations instead have <energy> $3J$. Thus exactly one of a minimum-energy triangle's bonds is frustrated. The bridge can simultaneously minimize its <energy> at $-J$ by requiring $t=-s'$, where $s'$ is the next cell's first base <spin>.
For fixed $s$ and $s'$, the number of allowed apex choices is therefore two when $s=s'$ and one when $s=-s'$. The zero-temperature counting <transfer matrix> on successive first base spins is
$$
G=\begin{pmatrix}2&1\\1&2\end{pmatrix},\qquad
\boxed{g_M=\operatorname{Tr}G^M=3^M+1,\qquad E_{\mathrm{GS}}=-2JM=-JL.}
$$
The <eigenvalues> of $G$ are $3$ and $1$. Dividing $k_B\ln g_M$ by the $3M$ sites and taking the <thermodynamic limit> gives $k_B\ln3/3$, independently confirming the derivative calculation. This does not assert a unique ordered <ground state>: the extensive choice of frustrated bonds is precisely what leaves <entropy> at zero <temperature>. At high <temperature>, all <Ising spins> become independent and $s\to k_B\ln2$ per site.
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