Can we have a vector field $\vec{V}(\vec{r})$ such that $\nabla\times\vec{V}=\vec{0}$ and $\nabla\cdot\vec{V}=0$
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$\begingroup$Cheap answer: sure just take a constant vector field so that all derivatives are zero.
A more interesting answer: a vector field in the plane which is both solenoidal and irrotational is basically the same thing as a holomorphic function in the complex plane. See here for more information on that.
So to get an example of an irrotational and solenoidal vector field in 3d space (which I guess is where you want to work judging from your cross product notation above), you can take any holomorphic function $f$ on $\mathbb{C}$ and form its "Polya vector field", basically $\overline f$ viewed as a vector field on $\mathbb{R}^2$. Then copy that vector field on every horizontal plane in 3d space.
Example: $f(z)=z^2$ is a holomorphic function. In terms of standard $(x,y)$ coordinates, this is $(x,y) \mapsto (x^2 -y^2 ,2xy)$. The associated Polya vector field is $(x,y) \mapsto (x^2-y^2,-2xy)$. Now make it a vector field on $\mathbb{R}^3$ by making it independent of $z$ and zeroing the third coordinate:$$\vec V(x,y,z) = (x^2-y^2,-2xy,0).$$
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