The divergence of the vector field \[{\rm{3xz\hat i}} + 2{\rm{xy\hat j}} – {\rm{y}}{{\rm{z}}^2}{\rm{\hat k}}\] at a point (1, 1, 1) is equal to A. 7 B. 4 C. 3 D. 0

$$\mathbf{F}(x, y, z) = 3xz\hat{\imath} + 2xy\hat{\jmath} – yz^2\hat{k}$$

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 $(1, 1, 1)$.