1. **Problem Statement:** Evaluate the line integral of the function $f(x,y,z) = x - 3y^2 + z$ over the line segment $C$ joining the points $O(0,0,0)$ to $A(1,1,1)$. 2. **Parametrization of the line segment $C$:** The line segment from $O$ to $A$ can be parametrized as: $$\vec{r}(t) = (1-0)t, (1-0)t, (1-0)t = (t, t, t)$$ where $t$ varies from $0$ to $1$. 3. **Explanation of parameter limits:** Since $t$ represents the fraction along the vector from $O$ to $A$, when $t=0$, the point is at $O$, and when $t=1$, the point is at $A$. Thus, the limits of $t$ are $0 \leq t \leq 1$. 4. **Express $f$ in terms of $t$:** Substitute $x = t$, $y = t$, and $z = t$ into $f$: $$f(t) = t - 3t^2 + t = 2t - 3t^2$$ 5. **Calculate $d\vec{r}$:** The derivative of $\vec{r}(t)$ is: $$\frac{d\vec{r}}{dt} = (1, 1, 1)$$ The magnitude is: $$|d\vec{r}| = \sqrt{1^2 + 1^2 + 1^2} dt = \sqrt{3} dt$$ 6. **Set up the line integral:** The line integral over $C$ is: $$\int_C f(x,y,z) \, ds = \int_0^1 f(t) |d\vec{r}| = \int_0^1 (2t - 3t^2) \sqrt{3} \, dt$$ 7. **Evaluate the integral:** $$\sqrt{3} \int_0^1 (2t - 3t^2) dt = \sqrt{3} \left[ t^2 - t^3 \right]_0^1 = \sqrt{3} (1 - 1) = 0$$ **Final answer:** $$\boxed{0}$$ **Summary:** The parameter $t$ runs from 0 to 1 because it represents the position along the line segment from $O$ to $A$. The line integral evaluates to 0.