Answer
$a = - \frac{{10}}{3}$
Work Step by Step
The two lines intersect if there exist parameter values ${t_1}$ and ${t_2}$ such that ${{\bf{r}}_1}\left( {{t_1}} \right) = {{\bf{r}}_2}\left( {{t_2}} \right)$, that is, if
$\left( {1,2,1} \right) + {t_1}\left( {1, - 1,1} \right) = \left( {3, - 1,1} \right) + {t_2}\left( {a,4, - 2} \right)$
In components we have
$x = 1 + {t_1} = 3 + a{t_2}$
$y = 2 - {t_1} = - 1 + 4{t_2}$
$z = 1 + {t_1} = 1 - 2{t_2}$
From the second equation we get ${t_2} = \frac{1}{4}\left( {3 - {t_1}} \right)$. Substituting it in the third equation gives ${t_1} = - 3$ and ${t_2} = \frac{3}{2}$.
Substituting ${t_1} = - 3$ and ${t_2} = \frac{3}{2}$ in the first equation, we obtain $a = - \frac{{10}}{3}$.
