The correct answer is $\boxed{\text{(B)}}$.
The directional derivative of a function $\phi$ at a point $P$ in the direction of a unit vector $\hat{u}$ is given by:
$$D_{\hat{u}} \phi(P) = \nabla \phi(P) \cdot \hat{u}$$
where $\nabla \phi(P)$ is the gradient of $\phi$ at $P$.
In this case, we have:
$$\phi(x, y, z) = 2xz – y^2$$
$$\nabla \phi(x, y, z) = (2x + 2z) \hat{i} + (2y) \hat{j} – 2y \hat{k}$$
Substituting $x = 1$, $y = 3$, and $z = 2$, we get:
$$\nabla \phi(1, 3, 2) = (4 + 4) \hat{i} + (6) \hat{j} – (4) \hat{k} = 4 \hat{i} – 6 \hat{j} + 2 \hat{k}$$
The directional derivative of $\phi$ at the point $(1, 3, 2)$ in the direction of $\hat{u}$ is then:
$$D_{\hat{u}} \phi(1, 3, 2) = (4 \hat{i} – 6 \hat{j} + 2 \hat{k}) \cdot \hat{u}$$
The maximum value of the directional derivative occurs when $\hat{u}$ is in the same direction as $\nabla \phi(1, 3, 2)$. Therefore, the maximum value of the directional derivative is:
$$\max_{\hat{u}} D_{\hat{u}} \phi(1, 3, 2) = (4 \hat{i} – 6 \hat{j} + 2 \hat{k}) \cdot (4 \hat{i} – 6 \hat{j} + 2 \hat{k}) = 24$$
Therefore, the directional derivative of $\phi$ at the point $(1, 3, 2)$ becomes maximum in the direction of $\boxed{\text{(B)}}$.