Answer
$$L = 5\sqrt {34} $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {2 + 3t,1 - 4t, - 4 + 3t} \right\rangle ,\,\,\,\,\,{\text{for }}1 \leqslant t \leqslant 6 \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {2 + 3t,1 - 4t, - 4 + 3t} \right\rangle \cr
& {\text{differentiating}} \cr
& {\bf{r}}'\left( t \right) = \left\langle {3, - 4,3} \right\rangle \cr
& {\text{use the Definition of Arc Length for Vector Functions }}\left( {{\text{see page 832}}} \right) \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& \,{\text{for }}1 \leqslant t \leqslant 6 \to a = 1{\text{ and }}b = 6.{\text{ then}} \cr
& L = \int_1^6 {\left| {\left\langle {3, - 4,3} \right\rangle } \right|} dt \cr
& L = \int_1^6 {\sqrt {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 3 \right)}^2}} } dt \cr
& {\text{simplifying}} \cr
& L = \int_1^6 {\sqrt {9 + 16 + 9} } dt \cr
& L = \int_1^6 {\sqrt {34} } dt \cr
& {\text{integrating}} \cr
& L = \sqrt {34} \left( t \right)_1^6 \cr
& L = \sqrt {34} \left( {6 - 1} \right) \cr
& L = 5\sqrt {34} \cr} $$
