Answer
${\vec{\triangledown}}\cdot ({\vec{\triangledown}\times \vec{v_a}}) = 0$
Work Step by Step
Consider a vector function $\vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}$
To show $\vec{\triangledown}\cdot (\vec{\triangledown} \times \vec{F}) = 0$
Consider,
$\vec{\triangledown} \times \vec{F} =
\begin{vmatrix}
\hat{i} && \hat{j} && \hat{k} \\
{\partial \over \partial x} && {\partial \over \partial y} && {\partial \over \partial z}\\
F_x && F_y && F_z
\end{vmatrix}$
$\vec{\triangledown} \times \vec{F} =
\hat{i}\big( {\partial \over \partial y} F_z - {\partial \over \partial z} F_y\big) + \hat{j}\big( {\partial \over \partial z} F_x - {\partial \over \partial x} F_z\big)+
\hat{k}\big( {\partial \over \partial x} F_y - {\partial \over \partial y} F_x\big) $
Now,
$\vec{\triangledown}\cdot (\vec{\triangledown} \times \vec{F}) =
{\partial \over \partial x}\big( {\partial \over \partial y} F_z - {\partial \over \partial z} F_y\big) + {\partial \over \partial y}\big( {\partial \over \partial z} F_x - {\partial \over \partial x} F_z\big) + {\partial \over \partial z}\big( {\partial \over \partial x} F_y - {\partial \over \partial y} F_x\big) = 0$
For $\vec{v_a} = x^2 \hat{i} + 3xz^2 \hat{j} − 2xz\hat{k}$
$\vec{\triangledown} \times \vec{v_a} = -6xz \hat{i} + 2z\hat{j} + 3z^2\hat{k}$
${\vec{\triangledown}}\cdot ({\vec{\triangledown}\times \vec{v_a}}) = -6z + 0 + 6z = 0$
Hence. Divergence of curl is always zero.
