= Solution
Write $\boldsymbol\Delta=\boldsymbol d-\boldsymbol b$. At an intersection, $\boldsymbol\Delta=\lambda\boldsymbol a-\mu\boldsymbol c$, so its <scalar triple product> with $\boldsymbol a,\boldsymbol c$ vanishes. The expression in the question is exactly
$$
\boldsymbol a\cdot(\boldsymbol c\times\boldsymbol d)+\boldsymbol c\cdot(\boldsymbol a\times\boldsymbol b)
=\boldsymbol\Delta\cdot(\boldsymbol a\times\boldsymbol c)=0.
$$
Under a shift of origin by $\boldsymbol h$, the direction vectors $\boldsymbol a,\boldsymbol c$ stay fixed, while $\boldsymbol b,\boldsymbol d$ both change by $-\boldsymbol h$. Thus $\boldsymbol\Delta$ 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 $\boldsymbol x_2-\boldsymbol x_1=\delta\lambda\boldsymbol a$ and $\boldsymbol x_4-\boldsymbol x_1=(\boldsymbol x_3-\boldsymbol x_1)+\delta\mu\boldsymbol c$. Bilinearity of the <cross product> gives
$$
\begin{aligned}
[(\boldsymbol x_3-\boldsymbol x_1)\times(\boldsymbol x_4-\boldsymbol x_1)]\cdot(\boldsymbol x_2-\boldsymbol x_1)
&=\delta\mu\,\delta\lambda\,[(\boldsymbol\Delta+\mu_1\boldsymbol c-\lambda_1\boldsymbol a)\times\boldsymbol c]\cdot\boldsymbol a\\
&=-\delta\mu\,\delta\lambda\,\boldsymbol\Delta\cdot(\boldsymbol a\times\boldsymbol c)\\
&=\boldsymbol\Delta\cdot[(\boldsymbol x_4-\boldsymbol x_3)\times(\boldsymbol x_2-\boldsymbol x_1)].
\end{aligned}
$$
The <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>.
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