Answer
\begin{align} r(t)& = \left\langle \frac{1}{3} t^3-\frac{1}{3}, \frac{5}{2} t^2-\frac{3}{2},t+1 \right\rangle . \end{align}
Work Step by Step
By integration, we have
\begin{align} r(t)& =\int r'(t) dt\\
& =\int \left\langle t^2,5t,1 \right\rangle dt\\& = \left\langle \frac{1}{3} t^3+c_1, \frac{5}{2} t^2+c_2,t+c_3\right\rangle .
\end{align} By the condition $r(1)=\lt 0,1,2\gt$, we get $$-\frac{1}{3}=c_1, \quad -\frac{3}{2}=c_2, \quad 1=c_3$$
Hence, we have \begin{align} r(t)& = \left\langle \frac{1}{3} t^3-\frac{1}{3}, \frac{5}{2} t^2-\frac{3}{2},t+1 \right\rangle . \end{align}
