I'll prove that they're neither A nor B, and therefore they must be C.

Part A: The lines are not parallel to each other
L1 is parallel to the vector (1, 3, -1), by reading off the "slopes" of the parameterization. L2 is parallel to the vector (2, 1, 4). Since (1,3,-1) is not equal to k * (2,1,4), they can't be parallel.

Part B: The lines have no intersection
If we set the x coordinates equal, we get 1 + t = 2s.
If we set the y coordinates equal, we get -2 + 3t = 3 + s
Solve that system yourself, finding t and s.
This tells you that there's some point where both the x coordinate AND the y coordinate of both lines are equal.
Check the z coordinate of each line. They should be different.

(Think of it this way...you're solving for x and y, which tells you that you're at the same point on a flat map. But when you plug in for z, you find that one line might be above ground, while the other point may be at a different altitude, or even below ground.)

View the original article here