Considering the straight line $$r:\left\{\begin{array}{rcl} 2x-y+z-2&=&0 \\ x+y+2z-7&=&0\end{array}\right.$$ determine its relative position to the straight line $$s: (x, y, z) = (1, 2, 3) + k \cdot (-1, 2, 0)$$.
Development:
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.
Solution:
The straight lines $$r$$ and $$s$$ cross.