Answer
Both ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ pass through $\left( {3, - 1,4} \right)$ and their direction vectors are parallel. Hence, they define the same line.
Work Step by Step
Let ${\bf{v}} = \left( {8,12, - 6} \right)$ and ${\bf{w}} = \left( {4,6, - 3} \right)$ be the direction vectors for ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$, respectively. Since ${\bf{v}} = 2{\bf{w}}$, 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 ${{\bf{r}}_1}\left( t \right)$, for instance, $P = \left( {3, - 1,4} \right)$ which corresponds to $t=0$. This point lies on ${{\bf{r}}_2}\left( t \right)$ if there is a value of $t$ such that
$\left( {3, - 1,4} \right) = \left( {11,11, - 2} \right) + t\left( {4,6, - 3} \right)$
This gives the equations
$3=11+4t$, ${\ \ }$ $-1=11+6t$, ${\ \ }$ $4=-2-3t$.
From the first equation we obtain $t=-2$. This value satisfies the second and the third equation. So, ${{\bf{r}}_2}\left( t \right)$ passes through $\left( {3, - 1,4} \right)$. Therefore, we conclude that ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ define the same line.
