 Solution

Write $\mathbf{\Delta }=\mathbf{d}-\mathbf{b}$. At an intersection, $\mathbf{\Delta }=\lambda \mathbf{a}-\mu \mathbf{c}$, so its scalar triple product with $\mathbf{a},\mathbf{c}$ vanishes. The expression in the question is exactlyUnder a shift of origin by $\mathbf{h}$, the direction vectors $\mathbf{a},\mathbf{c}$ stay fixed, while $\mathbf{b},\mathbf{d}$ both change by $-\mathbf{h}$. Thus $\mathbf{\Delta }$ and the condition are invariant. For nonparallel lines this is also sufficient for intersection; for parallel lines it only asserts coplanarity and need not imply intersection.

Put $\delta \lambda ={\lambda }_2-{\lambda }_1$ and $\delta \mu ={\mu }_2-{\mu }_1$. Then ${\mathbf{x}}_2-{\mathbf{x}}_1=\delta \lambda \mathbf{a}$ and ${\mathbf{x}}_4-{\mathbf{x}}_1=({\mathbf{x}}_3-{\mathbf{x}}_1)+\delta \mu \mathbf{c}$. Bilinearity of the cross product givesThe tetrahedron volume is one sixth of the absolute value of this triple product. Sliding either vertex pair along its line preserves $|\delta \lambda |$ or $|\delta \mu |$, because its separation is fixed and the direction vectors are fixed. Thus the volume is unchanged, including the zero-volume coplanar case. This is tetrahedron volume under sliding opposite edges.