The correct answer is $\boxed{\rm{\hat i} – 4{\rm{\hat j}}}$.
The vorticity vector is defined as the curl of the velocity vector, which is given by:
$$\omega = \nabla \times \overrightarrow V$$
In Cartesian coordinates, the curl can be written as:
$$\omega = \left( \dfrac{\partial Q}{\partial x} – \dfrac{\partial P}{\partial y} \right) {\rm{\hat i}} + \left( \dfrac{\partial P}{\partial z} – \dfrac{\partial Q}{\partial x} \right) {\rm{\hat j}} + \left( \dfrac{\partial R}{\partial y} – \dfrac{\partial Q}{\partial z} \right) {\rm{\hat k}}$$
where $P$, $Q$, and $R$ are the $x$, $y$, and $z$ components of the velocity vector, respectively.
In this case, the velocity vector is given by:
$$\overrightarrow V = 2xy\hat i – {{\rm{x}}^2}{\rm{z\hat j}}$$
Therefore, the components of the vorticity vector are:
$$\begin{align}
\dfrac{\partial Q}{\partial x} – \dfrac{\partial P}{\partial y} &= 2y – 0 = 2y \
\dfrac{\partial P}{\partial z} – \dfrac{\partial Q}{\partial x} &= -2xz = -2z \
\dfrac{\partial R}{\partial y} – \dfrac{\partial Q}{\partial z} &= 0 – 0 = 0
\end{align}$$
Therefore, the vorticity vector is given by:
$$\omega = 2y{\rm{\hat i}} – 2z{\rm{\hat j}}$$
Evaluating the vorticity vector at the point $(1, 1, 1)$ gives:
$$\omega = (2)(1){\rm{\hat i}} – (2)(1){\rm{\hat j}} = {\rm{\hat i}} – 2{\rm{\hat j}}$$