Answer
$$s = 3\sqrt {29} $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = - 3t{\bf{i}} + 2t{\bf{j}} + 4t{\bf{k}},{\text{ }}\left[ {0,3} \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ { - 3t{\bf{i}} + 2t{\bf{j}} + 4t{\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = - 3{\bf{i}} + 2{\bf{j}} + 4{\bf{k}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 4 \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {9 + 4 + 16} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {29} \cr
& {\text{Find the arc length }}s{\text{, using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^3 {\sqrt {29} } dt \cr
& {\text{Integrate}} \cr
& s = \sqrt {29} \left[ t \right]_0^3 \cr
& s = 3\sqrt {29} \cr
& \cr
& {\text{Graph}} \cr} $$
