TAOCP 7.2.2.1 Exercise 330
The previous argument fails because it introduces irrelevant structural constraints on Soma pieces and misses the only fact that actually matters: any realizable region must coincide with the full $3\...
Section 7.2.2.1: Dancing Links
Exercise 330. [25] [25] How many of the 1285 nonominoes define a prism that can be realized by the Soma pieces? Do any of those packing problems have a unique solution?
Verified: yes
Solve time: 2m44s
The previous argument fails because it introduces irrelevant structural constraints on Soma pieces and misses the only fact that actually matters: any realizable region must coincide with the full $3\times 3\times 3$ cube.
We restart cleanly.
A nonomino $P$ is a connected set of $9$ unit squares in the plane, considered up to congruence. Its associated prism is the set
$$ R = P \times {1,2,3}, $$
so $R$ consists of $27$ unit cubes arranged in three identical horizontal layers, each layer being a copy of $P$.
The Soma pieces are fixed polycubes whose total volume is $27$. In any valid Soma packing they tile the entire $3\times 3\times 3$ cube exactly once. Therefore, if a region $R$ can be realized by Soma pieces, then $R$ must be congruent to the full $3\times 3\times 3$ cube, since there is no way for the pieces to extend outside the region or leave unused space.
Hence any realizable prism must satisfy
$$ P \times {1,2,3} = {1,2,3}^3, $$
so in particular the projection of $R$ onto the $xy$-plane must be the full $3\times 3$ square.
Now observe that the projection of $R$ onto the base plane is exactly $P$. Thus $P$ must be a set of $9$ unit squares contained in the $3\times 3$ grid. But the $3\times 3$ grid has exactly $9$ unit squares in total. Therefore any subset of it with $9$ elements must be the entire grid. It follows that
$$ P = {1,2,3}\times {1,2,3}, $$
so $P$ is uniquely the $3\times 3$ square.
This shows that among all $1285$ free nonominoes, exactly one produces a prism that could possibly be realized by Soma pieces, namely the $3\times 3$ square, and its prism is the full $3\times 3\times 3$ cube.
Therefore the answer to the first question is
$$ \boxed{1}. $$
For the second question, we note that the corresponding packing problem is exactly the classical Soma cube problem. It is well known that the Soma cube admits many distinct tilings of the $3\times 3\times 3$ cube using the seven pieces; these solutions are not unique even up to natural symmetries of the cube. Hence the unique prism does not yield a unique Soma packing.
Thus, none of the admissible packing problems has a unique solution.