Correlation between parallel lines and volumes of tetrahedrons. Given 4 parallel lines $d_1$, $d_2$, $d_3$, $d_4$ , no more than 2 of which can be on a same plane. Plane (P) intersects the 4 lines at 4 points A, B, C, D. Plane (Q) ( not identical to plane (P)) intersects the 4 lines at $A_1$, $B_1$, $C_1$, $D_1$. Proof that the volumes of the 2 tetrahedra ABC
$D_1$ and $A_1$$B_1$$C_1$$D$ are equal.    
 A: Let $(P)||(Q).$
Thus, we need to prove that areas of $\Delta ABC$ and $\Delta A_1B_1C_1$ they are equal, 
which is obviously wrong: move $(Q)$ such that $(Q)||(P).$
A: It is not explicitly stated that one of the given four parallel lines passes through both $D$ and $D_1,$ but I will assume this is given since otherwise
the two tetrahedra might have different volumes.

If you know that a shear transformation of three-dimensional Euclidean space preserves volume, you can construct a plane $P'$ perpendicular to $d_1$
(hence also perpendicular to the other three given lines)
and apply a shear transformation parallel to $d_1$ that maps $P$ to $P'.$
This transformation maps $A,$ $B,$ and $C$ to points $A',$ $B',$ and $C'$
in plane $P'$, and the tetrahedron $ABCD_1$ has volume equal to $A'B'C'D_1,$
which has base $\triangle A'B'C'$ and height $DD_1.$
Likewise you can construct a plane $Q'$ perpendicular to $d_1$ and apply a shear transformation parallel to $d_1$ that maps $Q$ to $Q',$
and in particular maps $A_1,$ $B_1,$ and $C_1$ to $A_1',$ $B_1',$ and $C_1'.$
Then the tetrahedron $A_1B_1C_1D$ has volume equal to $A_1'B_1'C_1'D,$
which has base $\triangle A_1'B_1'C_1'$ and height $DD_1.$
But since $\triangle A_1'B_1'C_1'$ is congruent to $\triangle A'B'C',$
the tetrahedron $\triangle A_1'B_1'C_1'D$ has the same volume as $A'B'C'D_1,$
and hence all four tetrahedra have the same volume.
