Answer
\begin{align}
r(t)&= \left\langle \frac{1}{4} e^{2t-2}+\frac{3}{2}t-\frac{7}{4}, \frac{1}{12}t^4-\frac{1}{2}t^2+ \frac{2}{3}t -\frac{1}{4}, \frac{1}{2}t^2 -t+\frac{3}{2}\right\rangle .
\end{align}
Work Step by Step
By integration, we have
\begin{align}
r'(t)& = \left\langle \frac{1}{2} e^{2t-2}+c_1, \frac{1}{3}t^3-t+ c_2, t +c_3 \right\rangle .
\end{align}
By the condition $r'(1)=\lt 2,0,0\gt$, we get
$$\frac{3}{2}=c_1, \quad \frac{2}{3}=c_2, \quad -1=c_3$$
Hence we have
\begin{align}
r'(t)& = \left\langle \frac{1}{2} e^{2t-2}+\frac{3}{2}, \frac{1}{3}t^3-t+ \frac{2}{3}, t -1 \right\rangle .
\end{align}
Again, by integration we have
\begin{align}
r(t)& = \left\langle \frac{1}{4} e^{2t-2}+\frac{3}{2}t+a, \frac{1}{12}t^4-\frac{1}{2}t^2+ \frac{2}{3}t+b, \frac{1}{2}t^2 -t+c \right\rangle .
\end{align}
By the condition $r(1)=\lt0,0,1\gt$, we get
$$-\frac{7}{4}=a, \quad -\frac{1}{4}=b, \quad \frac{3}{2}=c$$
Hence we have
\begin{align}
r(t)&= \left\langle \frac{1}{4} e^{2t-2}+\frac{3}{2}t-\frac{7}{4}, \frac{1}{12}t^4-\frac{1}{2}t^2+ \frac{2}{3}t -\frac{1}{4}, \frac{1}{2}t^2 -t+\frac{3}{2}\right\rangle .
\end{align}
