Answer
Both ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ are parallel and they pass through a common point. Hence they define the same line.
Work Step by Step
Let ${\bf{v}} = \left( {2,1,3} \right)$ and ${\bf{w}} = \left( {8,4,12} \right)$ be the direction vectors for ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$, respectively. Since ${\bf{w}} = 4{\bf{v}}$, the two direction vectors are parallel. Therefore, the lines described by ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ are parallel. Next, we check if they have a point in common.
Let us choose any point on ${{\bf{r}}_2}\left( t \right)$, for instance, $P = \left( { - 6, - 3, - 9} \right)$ which corresponds to $t=0$. This point lies on ${{\bf{r}}_1}\left( t \right)$ if there is a value of $t$ such that
$\left( { - 6, - 3, - 9} \right) = t\left( {2,1,3} \right)$
This gives the equations
$-6=2t$, ${\ \ }$ $-3=t$, ${\ \ }$ $-9=3t$.
All three equations are satisfied with $t=-3$. So, ${{\bf{r}}_1}\left( t \right)$ passes through $P = \left( { - 6, - 3, - 9} \right)$. Therefore, we conclude that ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ define the same line.
