Find the value of integral $\int_{C}\left(z^{2}+y x\right) d s$, where $C$ is parmeterized by $\vec{r}(t)=\langle t,-3 t,-t\rangle$ for $0 \leq t \leq 5$

Step 1 ：Let the curve C be parameterized by \(\vec{r}(t)=\langle t,-3 t,-t\rangle\) for \(0 \leq t \leq 5\)

Step 2 ：Compute the derivative of \(\vec{r}(t)\) to get \(\vec{r}'(t) = \langle 1, -3, -1 \rangle\)

Step 3 ：Find the magnitude of \(\vec{r}'(t)\) to get \(||\vec{r}'(t)|| = \sqrt{11}\)

Step 4 ：Substitute \(\vec{r}(t)\) into the function \(f(z, y, x) = z^{2} + yx\) to get \(f(\vec{r}(t)) = -2t^{2}\)

Step 5 ：Substitute \(f(\vec{r}(t))\) and \(||\vec{r}'(t)||\) into the integral to get \(\int_{0}^{5} -2t^{2} \sqrt{11} dt\)

Step 6 ：Compute the integral to get \(-\frac{250\sqrt{11}}{3}\)

Step 7 ：Final Answer: \(\boxed{-\frac{250\sqrt{11}}{3}}\)