Find the distance between the lines $l_1:$ $x=1+4t,y=5-4t,z=-1+5t$ and $l_2:x=2+8t,y=4-3t,z=5+t$
So the approach in general is to find a vector that is orthogonal to 2 planes that the lines are in them.
I saw that the cross product is used to find this vector as in the following:
$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -4 & 5 \\ 8 & -3 & 1 \\ \end{vmatrix}=11\hat{i}+36\hat{j}+20\hat{k}$$
How can the cross product be used here? why does it give a vector that is orthogonal to 2 planes?
Because a line joining a point from one of the lines to a second point from the other line will be the shortest when it is perpendicular to both lines. Perpendicular to a line, means perpendicular to its generating vector, i.e. coefficients of its equation. And in three dimension all vectors that are simultaneously perpendicular to two vectors are multiples of their cross-product (when those vectors are not parallel.)