= Solution
At an intersection, $a+tu=b+t'v$, so $b-a=tu-t'v$. Alternating multilinearity of the <scalar triple product> then gives $[v,b-a,u]=t[v,u,u]-t'[v,v,u]=0$, establishing $[v,b,u]=[v,a,u]$.
For the location, write $b=a+tu-t'v$ and take its <scalar triple product> with $a$ and $v$:
$$
[a,b,v]=[a,a+tu-t'v,v]=t[a,u,v].
$$
Under the additional stated assumption $[a,u,v]\ne0$, division gives
$$
\boxed{r=a+\frac{[a,b,v]}{[a,u,v]}u.}
$$
These are <intersection parameters from scalar triple products>. The formula presupposes an intersection; the first coplanarity condition by itself also holds for distinct parallel lines. Intersections can exist with the displayed denominator zero, in which case a different auxiliary vector or a component calculation is needed.
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