Answer
The parametrizations are:
${{\bf{r}}_1}\left( t \right) = \left( {1 + 3t,4 + t,5 + 6t} \right)$
${{\bf{r}}_2}\left( t \right) = \left( {1 + 3t,t,6t} \right)$
Work Step by Step
Let ${{\bf{v}}_1}$ and ${{\bf{v}}_2}$ be the direction vectors of the lines parametrized by ${{\bf{r}}_1}$ and ${{\bf{r}}_2}$, respectively.
Since the parametrization ${{\bf{r}}_1}\left( t \right)$ of the line passing through $\left( {1,4,5} \right)$ and $\left( { - 2,3, - 1} \right)$, we have
${{\bf{v}}_1} = \left( {1,4,5} \right) - \left( { - 2,3, - 1} \right) = \left( {3,1,6} \right)$
We choose the point $\left( {1,4,5} \right)$ and using Eq. (5) of Section 13.2, we get the parametrization:
${{\bf{r}}_1}\left( t \right) = \left( {1,4,5} \right) + t\left( {3,1,6} \right)$
${{\bf{r}}_1}\left( t \right) = \left( {1 + 3t,4 + t,5 + 6t} \right)$
Since the parametrization ${{\bf{r}}_2}\left( t \right)$ of the second line is parallel to ${{\bf{r}}_1}\left( t \right)$, the two lines have the same direction vectors. Thus, we may choose ${{\bf{v}}_1} = {{\bf{v}}_2} = \left( {3,1,6} \right)$.
Since the second line passes through the point $\left( {1,0,0} \right)$, we use Eq. (5) of Section 13.2 to get the parametrization:
${{\bf{r}}_2}\left( t \right) = \left( {1,0,0} \right) + t\left( {3,1,6} \right)$
${{\bf{r}}_2}\left( t \right) = \left( {1 + 3t,t,6t} \right)$
