The answer is the ground state is 2-fold degenerate.
Instead of trying to construct Slater determinants, I will use a group theoretical argument based on permutations of 3 objects.
Suppose $E_a<E_b<E_c$ for the discussion.
The permutation group $S_3$ of $3$ objects has 3 irreducible representations, meaning $3$ types of sets of states with specified permutation symmetries, that will transform to a linear combination in the set under permutation. These irreps are labelled by partitions of $3$, and so are labelled as $(3),(2,1)$ and $(1,1,1)$.
The irrep $(3)$ is fully symmetric, and of dimension $1$. Thus, a possible $3$-particle spatial state of this type is $\Psi(x_1,x_2,x_3)\equiv \Psi(1,2,3)=\psi_a(1)\psi_a(2)\psi_a(3)$ for any single particle state $\psi_a(k)$ of particle $k$. Any permutation of the particles transforms $\Psi$ back to $+\Psi$.
For the spin part, there are in fact $4$ possible choices: they are
$\vert SM\rangle=\vert 3/2,M\rangle$ with
$\vert \frac32\frac32\rangle=\vert +_1\rangle\vert +_2\rangle\vert +_3\rangle\equiv \vert +++\rangle$. The other $3$ states are obtained from $\vert \frac32\frac32\rangle$ by lowering with $S_-=S_-^{(1)}+S_-^{(2)}+S_-^{(3)}$. Note that even if there are $4$ $S=3/2$ states, they are all clearly symmetric, i.e. any permutation of particle index will transform the state into itself. For instance: $P_{23}\vert +++\rangle=\vert +++\rangle$. Thus the permutations cannot change the values of $M$ and there is one symmetric state for each different value of $M$ with $S=\frac32$.
A basis for the $2$-dimensional irrep $(2,1)$ is a little harder to obtain but you can verify that
$$
\vert\psi_+\rangle=\frac{1}{\sqrt{2}}\vert aba\rangle
-\frac{1}{\sqrt{2}}\vert baa\rangle \\
\vert\psi_-\rangle=\frac{2}{\sqrt{6}}\vert aab\rangle
-\frac{1}{\sqrt{6}}\vert aba\rangle
-\frac{1}{\sqrt{6}}\vert baa \rangle
$$
transform into linear combinations of themselves under any permutation. Here:
$$
\vert aba\rangle\equiv \psi_a(1)\psi_b(2)\psi_a(3)\, , \quad \textit{etc}
$$
You can also verify in particular that
$$
P_{12}\vert\psi_\pm\rangle=\pm\vert\psi_\pm\rangle
$$
so these states are symmetric and antisymmetric under permutation of the first and second particles, but have more complicated symmetries if any other two indices are permuted.
The corresponding spin states are
$$
\vert\textstyle\frac12\frac12\rangle_+=\frac{1}{\sqrt{2}}\vert +-+\rangle
-\frac{1}{\sqrt{2}}\vert -++\rangle \\
\vert\textstyle\frac12\frac12\rangle_-=\frac{2}{\sqrt{6}}\vert ++-\rangle
-\frac{1}{\sqrt{6}}\vert +-+\rangle
-\frac{1}{\sqrt{6}}\vert -++ \rangle
$$
You can verify that $S_+\vert\textstyle\frac12\frac12\rangle_i=0$ and $S_z\vert\textstyle\frac12\frac12\rangle_k=\frac12\vert\textstyle\frac12\frac12\rangle_k$ so the states are legitimately labelled by $S=1/2$ and $M=1/2$.
Again: $P_{12}\vert\textstyle\frac12\frac12\rangle_\pm =
\pm \vert\textstyle\frac12\frac12\rangle_\pm$. Moreover, there are obviously two states with the same mixed symmetry, which transform into linear combo of themselves, of the type
$\vert\textstyle\frac12,-\frac12\rangle_\pm$ and obtained by lowering using $S_-$ on the $\vert\textstyle\frac12\frac12\rangle_\pm$ states.
Finally, there is the antisymmetric irrep $(1,1,1)$. A basis for this irrep is the determinant
$$
\phi(x_1,x_2,x_3)=\frac{1}{\sqrt{6}}
\left\vert\begin{array}{ccc}
\psi_a(x_1)&\psi_b(x_1)&\psi_c(x_1)\\
\psi_a(x_2)&\psi_b(x_3)&\psi_c(x_2)\\
\psi_a(x_3)&\psi_b(x_2)&\psi_c(x_3)
\end{array}\right\vert\, .
$$
There is no corresponding fully antisymmetric spin state as it is not possible to construct a $3\times 3$ determinant for three particles having only two possible spin orientations.
Now it is a result of group theory that, in order to obtain a state that is fully symmetric, one must combine:
- a fully symmetric spatial part with a fully antisymmetric spin part,
- a (2,1) mixed symmetry spatial part with a (2,1) mixed symmetry spin part,
- a fully antisymmetric spatial part with a fully symmetric spin part.
Option 1 is not possible because there is no fully antisymmetric spin state, as explained above.
Option 3 yields a spatial part which will have energy $E_a+E_b+E_c$, whereas option 2 gives a spatial part with energy $2E_a+E_b$. Thus, the ground state is obtained using option two.
We start by writing the most general combination
$$
\vert \Phi\rangle=\alpha \vert\psi_+\rangle\vert\textstyle\frac12\frac12\rangle_+
+\beta \vert\psi_-\rangle\vert\textstyle\frac12\frac12\rangle_+
+\gamma\vert\psi_+\rangle\vert\textstyle\frac12\frac12\rangle_-
+\delta\vert\psi_-\rangle\vert\textstyle\frac12\frac12\rangle_-
$$
For the state to be antisymmetric, we need $P_{12}\vert \Phi\rangle=-\vert \Phi\rangle$, which implies, using the result of $P_{12}$ on $\vert\psi_\pm\rangle$ and $\vert\textstyle\frac12\frac12\rangle_\pm$, that $\alpha=\delta=0$ so that
$$
\vert \Phi\rangle=
\beta \vert\psi_-\rangle\vert\textstyle\frac12\frac12\rangle_+
+\gamma\vert\psi_+\rangle\vert\textstyle\frac12\frac12\rangle_-
$$
Requiring now $P_{13}\vert\Phi\rangle=-\vert\Phi\rangle$ or $P_{23}\vert\Phi\rangle=-\vert\Phi\rangle$ fixes the relation between $\beta$ and $\gamma$ so there is only one state $\vert \Phi\rangle$ constructed from $\{\vert\psi_\pm\rangle\}$ and $\{\vert\frac12\frac12\rangle_\pm\}$ that is antisymmetric.
The second possible state is of the form
$$
\vert \Phi'\rangle=
\beta \vert\psi_-\rangle\vert\textstyle\frac12,-\frac12\rangle_+
+\gamma\vert\psi_+\rangle\vert\textstyle\frac12,-\frac12\rangle_-\, .
$$
Any other state of energy $2E_a+E_b$ will be a linear combination of $\vert \Phi\rangle$ and $\vert \Phi'\rangle$.
