(ii) $\quad \int_{C} \vec{F} \cdot d \vec{r}, \quad \vec{F}=x z \vec{\imath}+z^{3} \vec{\jmath}+y \vec{k}$ $C: \vec{r}(t)=e^{t} \vec{\imath}+e^{2 t} \vec{\jmath}+e^{-t} \vec{k}, \quad-1 \leq t \leq 1$

Step 1 ：Given the vector field \(\vec{F} = xz\vec{i} + z^{3}\vec{j} + y\vec{k}\) and the curve \(C\) parameterized by \(\vec{r}(t) = e^{t}\vec{i} + e^{2t}\vec{j} + e^{-t}\vec{k}\), with \(-1 \leq t \leq 1\).

Step 2 ：The line integral of the vector field \(\vec{F}\) along the curve \(C\) is given by \(\int_{C} \vec{F} \cdot d \vec{r} = \int_{a}^{b} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) dt\), where \(a\) and \(b\) are the limits of the parameter \(t\).

Step 3 ：First, we need to find \(\vec{r}'(t)\), the derivative of \(\vec{r}(t)\) with respect to \(t\).

Step 4 ：\(\vec{r}'(t) = e^{t}\vec{i} + 2e^{2t}\vec{j} - e^{-t}\vec{k}\)

Step 5 ：Then, we substitute \(\vec{r}(t)\) into \(\vec{F}\) to get \(\vec{F}(\vec{r}(t))\).

Step 6 ：\(\vec{F}(\vec{r}(t)) = e^{t}e^{-t}\vec{i} + (e^{-t})^{3}\vec{j} + e^{2t}\vec{k} = \vec{i} + e^{-3t}\vec{j} + e^{2t}\vec{k}\)

Step 7 ：Next, we compute the dot product \(\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t)\).

Step 8 ：\(\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) = e^{t} + 2e^{-t} - e^{t} = 2e^{-t}\)

Step 9 ：Finally, we integrate this from \(-1\) to \(1\).

Step 10 ：The integral is \(-2e^{-1} + 2e\)

Step 11 ：So, the line integral of the vector field \(\vec{F}\) along the curve \(C\) is \(\boxed{-2e^{-1} + 2e}\)