Answer
$$ \mathbf{r}=\left(-\frac{t^{2}}{2}+10\right) \mathbf{i}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{j}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{k}$$
Work Step by Step
Since
\begin{align*}
\frac{d \mathbf{r}}{d t}&=\int-(\mathbf{i}+\mathbf{j}+\mathbf{k}) d t\\
&=-(t \mathbf{i}+t \mathbf{j}+t \mathbf{k})+\mathbf{C}_{1}
\end{align*}
and
$$ \frac{d \mathbf{r}}{d t}(0)=\mathbf{0} \Rightarrow-(0 \mathbf{i}+0 \mathbf{j}+0 \mathbf{k})+\mathbf{C}_{1}=\mathbf{0} \Rightarrow \mathbf{C}_{1}=\mathbf{0} $$
Then
$$ \frac{d \mathbf{r}}{d t}=-(t \mathbf{i}+t \mathbf{j}+t \mathbf{k}) $$
and
\begin{align*}
\mathbf{r}&=\int-(t \mathbf{i}+t \mathbf{j}+t \mathbf{k}) d t\\
&=-\left(\frac{t^{2}}{2} \mathbf{i}+\frac{t}{2} \mathbf{j}+\frac{t^{2}}{2} \mathbf{k}\right)+\mathbf{C}_{2}
\end{align*}
$$ \mathbf{r}(0)=10 \mathbf{i}+10 \mathbf{j}+10 \mathbf{k}$$
Then
$$\mathbf{C}_{2}=10 \mathbf{i}+10 \mathbf{j}+10 \mathbf{k}$$
and
$$ \mathbf{r}=\left(-\frac{t^{2}}{2}+10\right) \mathbf{i}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{j}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{k}$$
