Answer:To show that $\overrightarrow{F} $ is solenoidal, we have to show that divergence of $\overrightarrow{F} $ is zero., i.e., $\nabla. \overrightarrow{F} =0.$

We know $\nabla=\bigg(\widehat{i}\dfrac{\partial}{\partial x}+\widehat{j}\dfrac{\partial}{\partial y}+\widehat{k}\dfrac{\partial}{\partial z}\bigg),$

so, $\bigg(\widehat{i}\dfrac{\partial}{\partial x}+\widehat{j}\dfrac{\partial}{\partial y}+\widehat{k}\dfrac{\partial}{\partial z}\bigg).[(y^2-z^2+3yz-2x) \widehat{i}+(3xz+2xy) \widehat{j}+(3xy-2xz+2z) \widehat{k}]$

$\Rightarrow \dfrac{\partial}{\partial x}(y^2-z^2+3yz-2x)+ \dfrac{\partial}{\partial y}(3xz+2xy)+ \dfrac{\partial}{\partial z} (3xy-2xz+2z)=-2+2x-2x+2=0.$

Hence vector $\overrightarrow{F} $ is solenoidal. 

To show that $\overrightarrow{F} $ is irrotational, we have to show that $curl \overrightarrow{F} =0$

$\Rightarrow \nabla \times\overrightarrow{F} =\begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y}& \dfrac{\partial}{\partial z} \\ y^2-z^2+3yz-2x & 3xz+2xy & 3xy-2xz+2z \end{vmatrix}\\=\widehat{i}(3x-3x)-\widehat{j}(3y-2z+2z-3y)+\widehat{k}(3z+2y-2y-3z)\\=0.$

Hence vector $\overrightarrow{F} $ is irrotational also.