Answer
The derivative at $t=3$:
$\frac{d}{{dt}}\left( {6{{\bf{r}}_1}\left( t \right) - 4{{\bf{r}}_2}\left( t \right)} \right){|_{t = 3}} = \left( {0, - 8, - 10} \right)$
Work Step by Step
$\frac{d}{{dt}}\left( {6{{\bf{r}}_1}\left( t \right) - 4{{\bf{r}}_2}\left( t \right)} \right) = 6{{\bf{r}}_1}'\left( t \right) - 4{{\bf{r}}_2}'\left( t \right)$
The derivative at $t=3$:
$\frac{d}{{dt}}\left( {6{{\bf{r}}_1}\left( t \right) - 4{{\bf{r}}_2}\left( t \right)} \right){|_{t = 3}} = 6{{\bf{r}}_1}'\left( 3 \right) - 4{{\bf{r}}_2}'\left( 3 \right)$
$ = 6\left( {0,0,1} \right) - 4\left( {0,2,4} \right)$
$\frac{d}{{dt}}\left( {6{{\bf{r}}_1}\left( t \right) - 4{{\bf{r}}_2}\left( t \right)} \right){|_{t = 3}} = \left( {0, - 8, - 10} \right)$
