Quantum linear optics Boson sampling

No and yes. The matrix $U$ will contain repeated columns (or rows depending on your you set it up) so that some simplifications are possible. Moreover, since your input state is in the irrep $(3,0,0)$ of $U(3)$, you can only get an output that is in this irrep. The irrep $(3,0,0)$ is of dimension $10$ and you can generate the basis states in the form begin{align} vert n_1 n_2 n_3rangle = frac{(a^dagger_1)^{n_1}(a^dagger_2)^{n_2}(a^dagger_3)^{n_3}}{sqrt{n_1!n_2!n_3!}}vert 0rangle, ,qquad n_1+n_2+n_3=3 end{align} so the output can only be a combination of those states.

The key point is that you cannot gets states in $(2,1,0)$ of $U(3)$ or in $(1,1,1)$ of $U(3)$ as they necessarily involve partially symmetric states, which need to be constructed using at least another degree of freedom to distinguish states in the same output ports. As an example one of the states in $(2,1,0)$ is of the form $$ leftvert begin{array}{cc} a^dagger_{1+}& a^dagger_{1-} a^dagger_{2+}& a^dagger_{2-}end{array}rightvert a^dagger_{1+} tag{1} $$ i.e. a determinant of creation operators multiplied by another creation operator. By antisymmetry of the determinant you see you need a second label (here $pm$) to distinguish different bosons but in your problems your bosons are indistinguishable so don't have such a second label, which eliminates are states of the form (1).

Thus in principle begin{align} Uvert 3,0,0rangle = sum_{n_1n_2n_3}vert n_1n_2n_3rangle U_{n_11}U_{n_21}U_{n_31}, .tag{2} end{align} Now if you detect bosons in some specific output channels - say $(a,b,c)$ then your detecting process is modelled as begin{align} Pi=frac{vert a,b,cranglelangle a,b,cvert}{a! b!c!} end{align} and thus only a limited number of terms in (2) will survive the projection associated with detection.