= Solution
A <boson> is an identical particle whose total <multiparticle quantum state> is symmetric under exchange of any two particles, whereas a <fermion> has a total state that is antisymmetric under every exchange. The <Spin-statistics theorem> associates integer spin with bosons and half-integer spin with fermions.
For three distinguishable spin-one particles, the spin <Hilbert space> has dimension
$$
3^3=27.
$$
Let
$$
\mathbf S=\mathbf S_1+\mathbf S_2+\mathbf S_3
$$
be the <total angular momentum operator>. Since each particle has
$$
\mathbf S_i^2=1(1+1)\hbar^2=2\hbar^2,
$$
we have
$$
\mathbf S^2=6\hbar^2
+2(\mathbf S_1\mathbin\cdot\mathbf S_2
+\mathbf S_2\mathbin\cdot\mathbf S_3
+\mathbf S_3\mathbin\cdot\mathbf S_1).
$$
The <Hamiltonian operator> is consequently
$$
H=\frac\lambda{\hbar^2}(\mathbf S^2-6\hbar^2).
$$
On the <total-spin sector> with quantum number $J$, its <energy eigenvalue> is
$$
E_J=\lambda[J(J+1)-6].
$$
The <Clebsch-Gordan decomposition> may be performed by first coupling particles $1$ and $2$. Their intermediate spin is $j_{12}=0,1,2$, and coupling the third spin gives
$$
\begin{array}{c|c}
j_{12}&J\\ \hline
0&1\\
1&0,1,2\\
2&1,2,3.
\end{array}
$$
Thus the <three spin-one angular-momentum decomposition> contains total spin $J=0,1,2,3$ with multiplicities $1,3,2,1$. Multiplying each multiplicity by the multiplet dimension $2J+1$ gives
$$
\begin{array}{c|c|c}
J&E_J&\text{degeneracy}\\ \hline
0&-6\lambda&1\\
1&-4\lambda&9\\
2&0&10\\
3&6\lambda&7.
\end{array}
$$
The degeneracies sum to $27$, as required.
When the particles are indistinguishable, their integer spin makes them bosons. Their common spatial wavefunction is symmetric, so their spin wavefunction must also lie in the <symmetric three-spin-one subspace>. Its decomposition is
$$
\operatorname{Sym}^3(1)=3\oplus1.
$$
Hence only the $J=3$ and $J=1$ levels remain, now with one copy of each multiplet:
$$
\boxed{
E=-4\lambda\text{ has degeneracy }3,
\qquad
E=6\lambda\text{ has degeneracy }7.}
$$
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