Evaluate $\int_{C} \vec{F} \cdot d \vec{r}$ where $\vec{F}=\langle 3 z, 4 y,-3 x\rangle$, and $C$ is given by $\vec{r}(t)=\langle t, \sin (t), \cos (t)\rangle, 0 \leq t \leq \pi$.

Step 1 ：Given the vector field \(\vec{F} = \langle 3z, 4y, -3x \rangle\) and the parameterized curve \(\vec{r}(t) = \langle t, \sin(t), \cos(t) \rangle\) where \(0 \leq t \leq \pi\).

Step 2 ：We are asked to evaluate the line integral \(\int_{C} \vec{F} \cdot d\vec{r}\).

Step 3 ：First, we substitute the components of \(\vec{r}(t)\) into \(\vec{F}\) to get a new vector function: \(\vec{F}(t) = \langle 3\cos(t), 4\sin(t), -3t \rangle\).

Step 4 ：Next, we find the derivative of \(\vec{r}(t)\), which is \(\vec{r}'(t) = \langle 1, \cos(t), -\sin(t) \rangle\).

Step 5 ：We then take the dot product of \(\vec{F}(t)\) and \(\vec{r}'(t)\) to get a scalar function: \(\vec{F}(t) \cdot \vec{r}'(t) = 3t\sin(t) + 4\sin(t)\cos(t) + 3\cos(t)\).

Step 6 ：Finally, we integrate this scalar function with respect to \(t\) from 0 to \(\pi\) to evaluate the line integral: \(\int_{0}^{\pi} \vec{F}(t) \cdot \vec{r}'(t) dt = 3\pi\).

Step 7 ：So, the value of the line integral is \(\boxed{3\pi}\).