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 $p2$.
HINT...find any vector joining one point on one line to another point on the other line and calculate the projection of this vector onto the common normal which you have found already.