Evaluate the line integral integral_C F * dr where F(x,y,z) = xz i - yz j + x k and C is the line segment from (-1, 2, 0) to (3, 0, 1).

The line segment goes from $P_0 = (-1, 2, 0)$ to $P_1 = (3, 0, 1)$.
A parameterization for a line segment is given by $\mathbf{r}(t) = P_0 + t(P_1 - P_0)$ for $0 \le t \le 1$.
First, find the vector $P_1 - P_0$:

$$P_1 - P_0 = (3 - (-1), 0 - 2, 1 - 0) = (4, -2, 1)$$

Now, write the parametric equations for $x, y, z$:

$$x(t) = -1 + 4t$$ $$y(t) = 2 - 2t$$ $$z(t) = t$$

Differentiate each component of $\mathbf{r}(t)$ with respect to $t$:

$$\frac{dx}{dt} = \frac{d}{dt}(-1 + 4t) = 4$$ $$\frac{dy}{dt} = \frac{d}{dt}(2 - 2t) = -2$$ $$\frac{dz}{dt} = \frac{d}{dt}(t) = 1$$

So, $d\mathbf{r} = \left( \frac{dx}{dt} \mathbf{i} + \frac{dy}{dt} \mathbf{j} + \frac{dz}{dt} \mathbf{k} \right) dt = (4 \mathbf{i} - 2 \mathbf{j} + 1 \mathbf{k}) dt$.

The given vector field is $\mathbf{F}(x,y,z) = xz \mathbf{i} - yz \mathbf{j} + x \mathbf{k}$.
Substitute $x(t)$, $y(t)$, and $z(t)$ into $\mathbf{F}$:

$$F_x = xz = (-1 + 4t)(t) = -t + 4t^2$$ $$F_y = -yz = -(2 - 2t)(t) = -(2t - 2t^2) = -2t + 2t^2$$ $$F_z = x = -1 + 4t$$

So, $\mathbf{F}(t) = (4t^2 - t) \mathbf{i} + (2t^2 - 2t) \mathbf{j} + (4t - 1) \mathbf{k}$.

$$\mathbf{F} \cdot d\mathbf{r} = (F_x \frac{dx}{dt} + F_y \frac{dy}{dt} + F_z \frac{dz}{dt}) dt$$ $$\mathbf{F} \cdot d\mathbf{r} = \left( (4t^2 - t)(4) + (2t^2 - 2t)(-2) + (4t - 1)(1) \right) dt$$ $$\mathbf{F} \cdot d\mathbf{r} = \left( (16t^2 - 4t) + (-4t^2 + 4t) + (4t - 1) \right) dt$$

Combine like terms:

$$\mathbf{F} \cdot d\mathbf{r} = (16t^2 - 4t^2 - 4t + 4t + 4t - 1) dt$$ $$\mathbf{F} \cdot d\mathbf{r} = (12t^2 + 4t - 1) dt$$

The integral is from $t=0$ to $t=1$:

$$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^1 (12t^2 + 4t - 1) dt$$

Integrate term by term:

$$= \left[ \frac{12t^3}{3} + \frac{4t^2}{2} - t \right]_0^1$$ $$= \left[ 4t^3 + 2t^2 - t \right]_0^1$$

Now, evaluate at the limits:

$$= \left( 4(1)^3 + 2(1)^2 - 1 \right) - \left( 4(0)^3 + 2(0)^2 - 0 \right)$$ $$= (4 + 2 - 1) - (0)$$ $$= 5$$

The value of the line integral is 5.