The divergence of a vector field is a measure of how much the vector field spreads out or converges at a point. It is calculated by taking the sum of the partial derivatives of the three components of the vector field with respect to their respective coordinates.
In this case, the vector field is given by
$$\mathbf{F}(x, y, z) = 3xz\hat{\imath} + 2xy\hat{\jmath} – yz^2\hat{k}$$
The divergence of $\mathbf{F}$ is then given by
$$\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x} (3xz) + \frac{\partial}{\partial y} (2xy) + \frac{\partial}{\partial z} \left(-yz^2\right) = 3x + 2y – 2yz$$
Evaluating this at the point $(1, 1, 1)$ gives
$$\nabla \cdot \mathbf{F}(1, 1, 1) = 3 + 2 – 2 = 3$$
Therefore, the divergence of the vector field $\mathbf{F}$ at the point $(1, 1, 1)$ is $\boxed{3}$.
The other options are incorrect because they do not represent the correct value of the divergence of the vector field $\mathbf{F}$ at the point $(1, 1, 1)$.