= Solution
For $\mathbf G$, the <product rule> for the <curl> gives
$$
\nabla\times\mathbf G
=\nabla(\mathbf a\cdot\mathbf r)\times\mathbf r
+(\mathbf a\cdot\mathbf r)(\nabla\times\mathbf r)
=\mathbf a\times\mathbf r.
$$
Since $\mathbf a\ne0$, this is not identically zero on $\mathbb R^3$. A <gradient field> has zero <curl>, so \b[no scalar potential exists for $\mathbf G$ on the whole space].
For the radial <vector field> $\mathbf K$, observe that
$$
\nabla\left[\frac12\log(1+x^2+y^2+z^2)\right]
=\frac{(x,y,z)^T}{1+x^2+y^2+z^2}=\mathbf K.
$$
Thus all <vector-field potentials> are
$$
\boxed{\phi(\mathbf r)=\frac12\log(1+|\mathbf r|^2)+C}.
$$
These are <smooth> everywhere, including the origin. Any two differ by a constant because their <gradient> difference vanishes on the <connected> whole space.
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