Calculate the distance between:
$$r:(x,y,z)=(2,1,3)+k\cdot(2,1,-1)$$
$$r':(x,y,z)=(-1,-1,4)+k\cdot(1,3,-2)$$
Development:
We start by determining the relative position of the straight lines.
First we verify that the governing vectors are not linearly dependent: $$$\left. \begin{array}{l} \vec{v}=(2,-1,1) \\ \vec{v}'=(1,3,-2) \end{array} \right\} \Rightarrow \dfrac{2}{1}\neq\dfrac{-1}{3}\neq\dfrac{1}{-2}$$$
The straight lines $$r$$ and $$r'$$ intersect or cross.
We take a point $$A$$ of $$r$$ and a point $$A'$$ of $$r'$$, and see if $$\{\overrightarrow{AA'},\vec{v},\vec{v}'\}$$ are linearly dependent or independent: $$$\left. \begin{array}{l} A = (2, 1, 3)\\ A' = (-1, -1, 4) \end{array} \right\} \Rightarrow \overrightarrow{AA'}=(-3,-2,1)$$$
$$$\begin{vmatrix} 2 & 1 & -3 \\ -1 & 3 & -2 \\ 1 & -2 & 1 \end{vmatrix} =0 \Rightarrow \text{rank}\big(\{\overrightarrow{AA'},\vec{v},\vec{v}'\}\big)=0$$$
Therefore the straight lines r and r' intersect and $$\text{d}(r,r') = 0$$.
Solution:
$$\text{d}(r,r') = 0$$