The scalar triple product is the determinant of the matrix with columns . A cyclic permutation of three columns is a product of two swaps, so its determinant sign is positive. Therefore . This uses the alternating multilinearity of the determinant, and remains true for linearly dependent vectors.
The determinant defining has two equal columns. Interchanging them both leaves it unchanged and reverses its sign, so it is zero. Equivalently, the cross product is orthogonal to , giving .
Three vectors in real three-dimensional space have zero determinant exactly when they are linearly dependent. Their linear span then has dimension at most two, so they lie in a common plane through the origin. Conversely vectors in such a plane span at most two dimensions, so their scalar triple product vanishes. This includes parallel nonzero vectors: they need not determine a unique plane, but they lie in at least one. Even zero vectors satisfy the determinant characterization, though the question excludes them.
By the vector triple product identity,
The last inner product vanishes by perpendicularity. Taking the inner product with gives
This is the identity for cross products with a common vector, and holds without nondegeneracy assumptions.
At an intersection, , so . Alternating multilinearity of the scalar triple product then gives , establishing .
For the location, write and take its scalar triple product with and :
Under the additional stated assumption , division gives
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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