Answer
The two lines intersect.
Intersection point: $\left( {3,4,7} \right)$.
Work Step by Step
Suppose the lines intersect and there is point of intersection at $t$ and $s$ such that ${{\bf{r}}_1}\left( t \right) = {{\bf{r}}_2}\left( s \right)$. So,
$\left( {0,1,1} \right) + t\left( {1,1,2} \right) = \left( {2,0,3} \right) + s\left( {1,4,4} \right)$
In component forms, we have
$x = t = 2 + s$, ${\ \ }$ $y = 1 + t = 4s$, ${\ \ }$ $z = 1 + 2t = 3 + 4s$
Solving the first two equations, we get $t=3$ and $s=1$. These values satisfies the third equation. Therefore, there exists point of intersection.
Substituting $t=3$ in ${{\bf{r}}_1}\left( t \right) = \left( {0,1,1} \right) + t\left( {1,1,2} \right)$ gives the intersection point: ${{\bf{r}}_1}\left( 3 \right) = \left( {3,4,7} \right)$.
Likewise, substituting $s=1$ in ${{\bf{r}}_2}\left( s \right) = \left( {2,0,3} \right) + s\left( {1,4,4} \right)$ also gives the same intersection point: ${{\bf{r}}_2}\left( 1 \right) = \left( {3,4,7} \right)$.
