Knowing \(\vrp'(t) = \la 2,\cos(t) , 12t\ra\text{,}\) we find \(\vrp(t)\) by evaluating the indefinite integral.
\begin{align*}
\int \vrp'(t)\,dt \amp = \la \int 2\,dt\,, \int \cos(t) \,dt\,, \int 12t\,dt\ra\\
\amp = \la 2t+C_1, \sin(t) + C_2, 6t^2 + C_3\ra\\
\amp = \la 2t,\sin(t) ,6t^2 \ra + \la C_1,C_2,C_3\ra\\
\amp = \la 2t,\sin(t) ,6t^2 \ra + \vec C\text{.}
\end{align*}
Note how each indefinite integral creates its own constant which we collect as one constant vector \(\vec C\text{.}\) Knowing \(\vrp(0) = \la 5,3,0\ra\) allows us to solve for \(\vec C\text{:}\)
\begin{align*}
\vrp(t) \amp = \la 2t,\sin(t) ,6t^2 \ra + \vec C\\
\vrp(0) \amp = \la 0,0,0 \ra + \vec C\\
\la 5,3,0\ra \amp = \vec C\text{.}
\end{align*}
So \(\vrp(t) = \la 2t,\sin(t) ,6t^2\ra + \la 5,3,0\ra = \la 2t+5, \sin(t) + 3, 6t^2\ra\text{.}\) To find \(\vec r(t)\text{,}\) we integrate once more.
\begin{align*}
\int \vrp(t)\,dt \amp = \la \int 2t+5\,dt, \int \sin(t) + 3\,dt, \int 6t^2\,dt \ra\\
\amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \vec C\text{.}
\end{align*}
With \(\vec r(0) = \la -7,-1,2\ra\text{,}\) we solve for \(\vec C\text{:}\)
\begin{align*}
\vec r(t) \amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \vec C\\
\vec r(0) \amp = \la 0,-1,0\ra + \vec C\\
\la -7,-1,2\ra \amp = \la 0,-1,0\ra + \vec C\\
\la -7,0,2\ra \amp = \vec C\text{.}
\end{align*}
So
\begin{align*}
\vec r(t) \amp = \la t^2+5t, -\cos(t) + 3t, 2t^3\ra + \la -7,0,2\ra\\
\amp = \la t^2+5t-7,-\cos(t) +3t,2t^3+2\ra\text{.}
\end{align*}