158 Let $A B C D$ be a non-degenerate tetrahedron (i.e., $A, B, C, D$ are not coplanar), then the number of different parallelepipeds in space with points $A, B, C, D$ as four of its vertices is ( )
A. 4
B. 5
C. 29
D. None of the above answers is correct
$158 \mathrm{C}$. Considering $A, B, C, D$ as four of the vertices to form a parallelepiped, we need to determine how many faces of the original tetrahedron are exposed.
(1) If the tetrahedron $\Lambda B C D$ has no faces exposed, then its two opposite edges must be the diagonals of two opposite faces of the parallelepiped. There is only 1 such parallelepiped.
(2) If the tetrahedron $A B C D$ has three faces exposed, then the three edges emanating from one of its vertices must be three adjacent edges of the parallelepiped. There are 4 such parallelepipeds (since there are 4 ways to choose this vertex).
(3) If the tetrahedron $A B C D$ has exactly two faces exposed, then the common edge of these two faces (which is also an edge of the parallelepiped) has 6 possible choices. Once this edge is chosen, one of the other two edges on the same face can be chosen as an edge of the parallelepiped (the other being the diagonal of the parallelepiped's face). Once an edge on the face is chosen, the parallelepiped is determined. Therefore, there are $6 \times 2 = 12$ such parallelepipeds.
(4) If the tetrahedron $A B C D$ has exactly one face exposed, then one of its edges must be the body diagonal of the parallelepiped. Fixing the exposed face of the tetrahedron, there are 3 ways to choose this edge (i.e., any one of the three edges from the vertex outside the face). Additionally, there are 4 ways to choose the exposed face, so there are 12 such tetrahedra.
In summary, the number of parallelepipeds with $A, B, C, D$ as four of the vertices is $1 + 4 + 12 + 12 = 29$.