= Solution
By <addition of angular momentum>, combining the first two <spin one-half> particles gives $0\oplus1$. Adding the third gives the <three spin-one-half angular-momentum decomposition>
$$
\left(\frac12\right)^{\otimes3}
=\left(0\oplus1\right)\otimes\frac12
=\frac12\oplus\frac12\oplus\frac32.
$$
Thus there is one $S=3/2$ <total-spin sector> and two independent $S=1/2$ sectors. Since the particles are distinguishable, no <fermionic exchange symmetry> restriction removes either copy. In a simultaneous <angular momentum eigenstate>, $\hat S^2$ and $\hat S^{(z)}$ have <eigenvalues> $\hbar^2S(S+1)$ and $\hbar M$, so
$$
\boxed{E_{S,M}=\epsilon_t S(S+1)+\epsilon_z M.}
$$
The <energy levels> and their generic multiplicities are
$$
\begin{array}{c|c|c|c}
S&M&E_{S,M}&\text{multiplicity}\\ \hline
3/2&-3/2,-1/2,1/2,3/2&15\epsilon_t/4+\epsilon_zM&1\text{ for each }M\\
1/2&-1/2,1/2&3\epsilon_t/4+\epsilon_zM&2\text{ for each }M
\end{array}
$$
The count $4+2\times2=8$ agrees with the <dimension> of the three-particle <tensor product>. These multiplicities must be added whenever numerical <energy eigenvalues> coincide. In particular, for $\epsilon_z=0$ and $\epsilon_t\ne0$, there are two fourfold <degenerate energy levels>, $15\epsilon_t/4$ and $3\epsilon_t/4$. For $\epsilon_t=0$ and $\epsilon_z\ne0$, the multiplicities at $M=-3/2,-1/2,1/2,3/2$ are $1,3,3,1$. If both coefficients vanish, the single <energy level> has multiplicity eight. Other coincidences between the two <total-spin sectors> occur when $3\epsilon_t+(M-m)\epsilon_z=0$, with $M$ from the quartet and $m=\pm1/2$ from a doublet.
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