Relative position of two straight lines

We start by calculating the relative position between two straight lines. We will do it in a geometric way. Thus we need the director vector of $r$.

Note that the implicit equations of the straight line in fact consist of two equations of two planes that cut, determining a straight line.

Therefore, we can obtain the director vector of $r$, $\overrightarrow{v}$ by doing the vector product between the normal vectors of the planes:

$$\overrightarrow{v}=\overrightarrow{n_1}\times\overrightarrow{n_2}=\left|\begin{matrix} i & j & k \\ 2 & -1 & 1 \\ 1 & 1 & 2 \end{matrix} \right|=-2i+j+2k+k-4j-i=-3i-3j+3k=(-3,-3,3)$$

We can take for simplicity, $\overrightarrow{v}=(1,1,-1)$, although we already observe that the straight lines are neither parallel nor the same since its director vectors $\overrightarrow{u}=(-1, 2, 0)$ and $\overrightarrow{v}=(1, 1,-1)$ are not parallel.

We look for points $A$ belonging to $r$, and $A'$ belonging to $s$: $$A=(0,1,3) \ \ ; \ \ A'=(1,2,3) \Rightarrow \overrightarrow{AA'}=(1,1,0)$$

We look finally to see if $\overrightarrow{AA'}$, $\overrightarrow{u}$ and the $\overrightarrow{v}$ are linearly dependent or independent: $$\left|\begin{matrix} 1 & 1 & 0 \\ -1 & 2 & 0 \\ 1 & 1 & -1 \end{matrix} \right|=-2-1=-3$$

Therefore the vectors are linearly independent and the straight lines cross.