= 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> gives
$$
[\nabla\times(\mathbf a\times\mathbf r)]_i
=\epsilon_{ijk}\epsilon_{klm}\partial_j(a_lx_m)=2a_i.
$$
Thus $\boxed{\nabla\times(\mathbf a\times\mathbf r)=2\mathbf a}$. A <conservative vector field> has zero <curl>, so the field is \b[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.
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