Answer: (i) \(\nabla\times\vec A=-3\hat i+0\hat j-3\hat k\) at \((1,2,-1)\); (ii) \(\nabla\times\vec B=\vec0\), so \(\vec B\) is irrotational.
\[\nabla\times\vec A=\begin{vmatrix}\hat i&\hat j&\hat k\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ x^2yz&xy^2z&xyz^2\end{vmatrix}\]
\(\hat i\): \(\dfrac{\partial}{\partial y}(xyz^2)-\dfrac{\partial}{\partial z}(xy^2z)=xz^2-xy^2\)
\(\hat j\): \(\dfrac{\partial}{\partial z}(x^2yz)-\dfrac{\partial}{\partial x}(xyz^2)=x^2y-yz^2\)
\(\hat k\): \(\dfrac{\partial}{\partial x}(xy^2z)-\dfrac{\partial}{\partial y}(x^2yz)=y^2z-x^2z\)
\[\nabla\times\vec A=(xz^2-xy^2)\hat i+(x^2y-yz^2)\hat j+(y^2z-x^2z)\hat k\]
At \((1,2,-1)\): \(\hat i: 1-4=-3\); \(\hat j: 2-2=0\); \(\hat k: (4)(-1)-(1)(-1)=-3\).
\[\nabla\times\vec A\big|_{(1,2,-1)}=-3\hat i-3\hat k\]
(ii) \(\nabla\times\vec B\):
\(\hat i\): \(\dfrac{\partial(xy)}{\partial y}-\dfrac{\partial(zx)}{\partial z}=x-x=0\); \(\hat j\): \(\dfrac{\partial(yz)}{\partial z}-\dfrac{\partial(xy)}{\partial x}=y-y=0\); \(\hat k\): \(\dfrac{\partial(zx)}{\partial x}-\dfrac{\partial(yz)}{\partial y}=z-z=0\).
So \(\nabla\times\vec B=\vec 0\) everywhere and \(\vec B\) is irrotational (in fact \(\vec B=\nabla(xyz)\)).
