Answer
$$\left\langle {\frac{{2{{\left( 6 \right)}^{5/2}}}}{5},\frac{{2{{\left( 6 \right)}^{5/2}}}}{5},6} \right\rangle $$
Work Step by Step
$$\eqalign{
& \int_{ - 3}^3 {\left\langle {{{\left( {3 - t} \right)}^{3/2}},{{\left( {3 + t} \right)}^{3/2}},1} \right\rangle } dt \cr
& {\text{Then}}{\text{,}} \cr
& = \left\langle {\int_{ - 3}^3 {{{\left( {3 - t} \right)}^{3/2}}dt} ,\int_{ - 3}^3 {{{\left( {3 + t} \right)}^{3/2}}} ,\int_{ - 3}^3 {dt} } \right\rangle \cr
& {\text{Integrating the components yields}} \cr
& = \left\langle {\left[ { - \frac{{{{\left( {3 - t} \right)}^{5/2}}}}{{5/2}}} \right]_{ - 3}^3,\left[ {\frac{{{{\left( {3 + t} \right)}^{5/2}}}}{{5/2}}} \right]_{ - 3}^3,\left[ t \right]_{ - 3}^3} \right\rangle \cr
& = \left\langle {\left[ { - \frac{{2{{\left( {3 - t} \right)}^{5/2}}}}{5}} \right]_{ - 3}^3,\left[ {\frac{{2{{\left( {3 + t} \right)}^{5/2}}}}{5}} \right]_{ - 3}^3,\left[ t \right]_{ - 3}^3} \right\rangle \cr
& {\text{Use the vector form of the Fundamental Theorem of Calculus}} \cr
& \int_a^b {{\bf{r}}\left( t \right)} dt = \left[ {{\bf{R}}\left( t \right)} \right]_a^b = {\bf{R}}\left( b \right) - {\bf{R}}\left( a \right) \cr
& {\text{then}} \cr
& = \left\langle { - \left[ {\frac{{2{{\left( {3 - 3} \right)}^{5/2}}}}{5} - \frac{{2{{\left( {3 + 3} \right)}^{5/2}}}}{5}} \right],\left[ {\frac{{2{{\left( {3 + 3} \right)}^{5/2}}}}{5} - \frac{{2{{\left( {3 - 3} \right)}^{5/2}}}}{5}} \right],\left[ {3 + 3} \right]} \right\rangle \cr
& {\text{simplify}} \cr
& = \left\langle { - \left[ { - \frac{{2{{\left( 6 \right)}^{5/2}}}}{5}} \right],\left[ {\frac{{2{{\left( 6 \right)}^{5/2}}}}{5}} \right],6} \right\rangle \cr
& = \left\langle {\frac{{2{{\left( 6 \right)}^{5/2}}}}{5},\frac{{2{{\left( 6 \right)}^{5/2}}}}{5},6} \right\rangle \cr} $$
