Answer
$$\mathbf{r}=\left(-\frac{t^{2}}{2}+1\right) \mathbf{i}+\left(-\frac{t^{2}}{2}+2\right) \mathbf{j}+\left(-\frac{t^{2}}{2}+3\right) \mathbf{k}$$
Work Step by Step
Given \begin{aligned} \frac{d \mathbf{r}}{d t} &=-t \mathbf{i}-t \mathbf{j}-t \mathbf{k} \\ \mathbf{r}(0) &=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k} \end{aligned}
Then
\begin{align*}
\mathbf{r}&=\int(-t \mathbf{i}-t \mathbf{j}-t \mathbf{k}) d t\\
&=-\frac{t^{2}}{2} \mathbf{i}-\frac{t^{2}}{2} \mathbf{j}-\frac{t^{2}}{2} \mathbf{k}+\mathbf{C}
\end{align*}
Since $\mathbf{r}(0) =\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}$, then
\begin{align*}
\mathbf{r}(0)&=0 \mathbf{i}-0 \mathbf{j}-0 \mathbf{k}+\mathbf{C}\\
&=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k} \\
\mathbf{C}&=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}
\end{align*}
Hence $$\mathbf{r}=\left(-\frac{t^{2}}{2}+1\right) \mathbf{i}+\left(-\frac{t^{2}}{2}+2\right) \mathbf{j}+\left(-\frac{t^{2}}{2}+3\right) \mathbf{k}$$
