Answer
\begin{align}
r(t)&= \left\langle 1, t^2-6t+10, t-3\right\rangle .
\end{align}
Work Step by Step
By integration, we have
\begin{align}
r'(t)& = \left\langle c_1, 2t+ c_2, c_3 \right\rangle .
\end{align}
By the condition $r'(3)=\lt 0,0,1\gt$, we get
$$0=c_1, \quad -6=c_2, \quad 1=c_3$$
Hence we have
\begin{align}
r'(t)& = \left\langle 0, 2t-6, 1 \right\rangle .
\end{align}
Again, by integration we have
\begin{align}
r(t)& = \left\langle a, t^2-6t+b, t+c \right\rangle .
\end{align}
By the condition $r(3)=\lt1,1,0\gt$, we get
$$1=a, \quad 10=b, \quad -3=c$$
Hence we have
\begin{align}
r(t)&= \left\langle 1, t^2-6t+10, t-3\right\rangle .
\end{align}
