I am tasked to find the distance between these two lines.
$p1 ... x = 1 + t, y = -1 + 2t, z = t$
$p2 ... x = 1 - t; y = 3 - t; z = t$
Those two lines are nonparallel and they do not intersect (I checked that).
Using the vector product I computed the normal (the line orthogonal to both of these lines), and the normal is $(3, -2, 1)$. Now I have the direction vector of the line which will intersect both of my nonparallel lines.
However, here's where I encounter the problem - I don't know what next. The next logical step in my opinion would be to find a point on $p1$ where I could draw that orthogonal line and where that orthogonal line would also intersect with $p2$... There's only one such point, since we are in 3D space and I could draw an orthogonal line from any point in $p1$ but it could miss $p2$.