Answer
Point of intersection: $\left( { - 2,1,0} \right)$
Work Step by Step
To find the point of intersection we must find ${t_1}$ and ${t_2}$ such that ${\bf{r}}\left( {{t_1}} \right) = {\bf{s}}\left( {{t_2}} \right)$. So,
$\left( {1,0,0} \right) + {t_1}\left( { - 3,1,0} \right) = \left( {0,1,1} \right) + {t_2}\left( {2,0,1} \right)$
In component forms, we have
$x = 1 - 3{t_1} = 2{t_2}$, ${\ \ }$ $y = {t_1} = 1$, ${\ \ }$ $z = 0 = 1 + {t_2}$
Solving the first two equations, we get ${t_1} = 1$ and ${t_2} = - 1$. These values satisfies the third equation. Therefore, there exists point of intersection.
Substituting ${t_1} = 1$ in ${\bf{r}}\left( t \right) = \left( {1,0,0} \right) + t\left( { - 3,1,0} \right)$ gives the intersection point: ${\bf{r}}\left( {{t_1}} \right) = \left( { - 2,1,0} \right)$
Likewise, substituting ${t_2} = - 1$ in ${\bf{s}}\left( t \right) = \left( {0,1,1} \right) + t\left( {2,0,1} \right)$ also gives the same intersection point: ${\bf{s}}\left( {{t_2}} \right) = \left( { - 2,1,0} \right)$
