Answer
a)$\vec{\nabla}\times\vec{v_a}=-6xz\hat{x}+2z\hat{y}+3z^2\hat{z}$
b)$\vec{\nabla}\times\vec{v_b}=-2y\hat{x}-3z\hat{y}-x\hat{z}$
c)$\vec{\nabla}\times\vec{v_c}=0$
Work Step by Step
a)$\vec{\nabla}\times\vec{v_a}=(\hat{x}\frac{\partial}{\partial{x}}+\hat{y}\frac{\partial}{\partial{y}}+\hat{z}\frac{\partial}{\partial{z}})\times(x^2\hat{x}+3xz^2\hat{y}−2xz\hat{z})$
$\hspace{1.8cm}=\hat{x}(0-6xz)+\hat{y}(0+2z)+\hat{z}(3z^2-0)$
$\hspace{1.8cm}=\hat{x}(-6xz)+\hat{y}(2z)+\hat{z}(3z^2)$
b)$\vec{\nabla}\times\vec{v_b}=(\hat{x}\frac{\partial}{\partial{x}}+\hat{y}\frac{\partial}{\partial{y}}+\hat{z}\frac{\partial}{\partial{z}})\times(xy\hat{x}+2yz\hat{y}+3xz\hat{z})$
$\hspace{1.8cm}=\hat{x}(0-2y)+\hat{y}(0-3z)+\hat{z}(0-x)$
$\hspace{1.8cm}=\hat{x}(-2y)+\hat{y}(-3z)+\hat{z}(-x)$
c)$\vec{\nabla}\times\vec{v_c}=(\hat{x}\frac{\partial}{\partial{x}}+\hat{y}\frac{\partial}{\partial{y}}+\hat{z}\frac{\partial}{\partial{z}})\times(y^2\hat{x}+(2xy+z^2)\hat{y}+2yz\hat{z})$
$\hspace{1.8cm}=\hat{x}(2z-2z)+\hat{y}(0-0)+\hat{z}(2y-2y)=0$
