 Solution

Since $\mathbf{a}$ is constant, direct differentiation gives $(\mathbf{r}\cdot \nabla )\mathbf{a}=0$, $(\mathbf{a}\cdot \nabla )\mathbf{r}=\mathbf{a}$, $\nabla \cdot \mathbf{r}=3$ and $\nabla \cdot \mathbf{a}=0$. The stated right-hand side is therefore $2\mathbf{a}$. Independently, index notation givesThus $\boxed{\nabla \times (\mathbf{a}\times \mathbf{r})=2\mathbf{a}}$. A conservative vector field has zero curl, so the field is not conservative unless $\mathbf{a}=0$. For a closed planar circuit, its circulation is $2\mathbf{a}\cdot \mathbf{A}$ by Stokes theorem; the zero-vector exception is conservative with a constant potential.