Answer
$$s = \sqrt {26} $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = - t{\bf{i}} + 4t{\bf{j}} + 3t{\bf{k}},{\text{ }}\left[ {0,1} \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ { - t{\bf{i}} + 4t{\bf{j}} + 3t{\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = - {\bf{i}} + 4{\bf{j}} + 3{\bf{k}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 3 \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {1 + 16 + 9} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {26} \cr
& {\text{Find the arc length }}s{\text{ using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^1 {\sqrt {26} } dt \cr
& {\text{Integrate}} \cr
& s = \sqrt {26} \left[ t \right]_0^1 \cr
& s = \sqrt {26} \cr
& \cr
& {\text{Graph}} \cr} $$
