Finding points on a line that are closest 
Find the points that give the shortest distance between the lines$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2-t\\-1+2t\\-1+t\end{pmatrix}\\\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}5+3s\\0\\2-s\end{pmatrix}$$

So I subtracted the second line from the first to get these two equations:
$\begin{pmatrix}-t-3s-3\\ -1+2t\\ t+s-3\end{pmatrix}\cdot \begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}=0$
$\begin{pmatrix}-t-3s-3\\ -1+2t\\ t+s-3\end{pmatrix}\cdot \begin{pmatrix}3\\ 0\\ -1\end{pmatrix}=0$
I know I am supposed to rearrange them together two get a system of equations but I am not sure how.
Any help please?

Two lines in $\mathbb R^3$ either intersect (distance = 0) or not intersect (distance = the distance between two parallel planes on which the two lines lie accordingly). Classify it first, then use $u\times v$, where $u, v$ indicate the direction of the two lines, to find the normal for the plane. Hope that helps!
