= Solution
Use the <Unscaled Papkovich–Neuber representation>
$$
\mathbf u=\nabla(\mathbf x\cdot\boldsymbol\Phi+\chi)-2\boldsymbol\Phi,\qquad p=2\mu\nabla\cdot\boldsymbol\Phi,
$$
with <harmonic> <vector> and <scalar> potentials. For a torque-driven <sphere> choose
$$
\boldsymbol\Phi=\frac{\mathbf G\times\nabla(1/r)}{16\pi\mu},\qquad\chi=0.
$$
Its components are <harmonic> for $r>a$, and both $\mathbf x\cdot\boldsymbol\Phi$ and $\nabla\cdot\boldsymbol\Phi$ vanish. Hence
$$
\boxed{\mathbf u_G=\frac{\mathbf G\times\mathbf x}{8\pi\mu r^3},\quad p_G=0,\quad\boldsymbol\Omega_G=\frac{\mathbf G}{8\pi\mu a^3}.}
$$
The last equality follows from the <no-slip boundary condition> at $r=a$. To check that $\mathbf G$ is indeed the applied <couple>, the <traction> on the <sphere>, with $\mathbf n=\mathbf x/a$ directed into the fluid, is $\boldsymbol\sigma_G\mathbf n=-3\mu\boldsymbol\Omega_G\times\mathbf n$. Its <hydrodynamic torque> is
$$
a\int_{r=a}\mathbf n\times(-3\mu\boldsymbol\Omega_G\times\mathbf n)dS=-8\pi\mu a^3\boldsymbol\Omega_G=-\mathbf G.
$$
Here $\int\mathbf n\mathbf n\,dS=(4\pi a^2/3)I$. The applied <couple> balances this resisting <torque>. The resulting field is the <rotlet> of a <rotating sphere in Stokes flow>.
For the <Lorentz reciprocal theorem for Stokes flow>, let $\mathbf n_D$ point out of the fluid domain $D$ and take the same <viscosity> in both flows. With $\nabla\cdot\boldsymbol\sigma_i+\mathbf f_i=0$, the theorem is
$$
\boxed{\int_{\partial D}(\mathbf u_1\cdot\boldsymbol\sigma_2\mathbf n_D-\mathbf u_2\cdot\boldsymbol\sigma_1\mathbf n_D)dS=\int_D(\mathbf u_2\cdot\mathbf f_1-\mathbf u_1\cdot\mathbf f_2)dV.}
$$
It follows by taking the <divergence> of the boundary integrand: <incompressibility> removes pressure-work terms, the two Newtonian strain contractions are equal, and the remaining <stress> <divergences> give the displayed body-force terms. This convention matters: on the inner boundary of an exterior domain, $\mathbf n_D=-\mathbf n$.
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