= Solution
Use the <free-draining approximation>: neglect the <hydrodynamic interaction> between the widely separated spheres and neglect the connector's viscous drag. <Stokes law> gives $\mathbf f_i=-6\pi\mu a_i\mathbf U$ for each sphere. Their resultant is
$$
\mathbf F=\mathbf f_1+\mathbf f_2=-6\pi\mu(a_1+a_2)\mathbf U,
\qquad\boxed{A\simeq6\pi\mu(a_1+a_2)I.}
$$
This is the leading two-sphere approximation, not an exact finite-separation <hydrodynamic resistance matrix>. Writing $d=|\mathbf r_2-\mathbf r_1|$ and $\mathbf n=(\mathbf r_2-\mathbf r_1)/d$, the first <Stokeslet> reflection gives, if the connector remains negligible,
$$
A=6\pi\mu(a_1+a_2)I-\frac{9\pi\mu a_1a_2}{d}(I+\mathbf n\mathbf n^T)+O\!\left(\mu\frac{(a_1+a_2)^3}{d^2}\right).
$$
The negative correction expresses reduced drag from fluid entrainment by the other sphere. A very thin but arbitrarily long rod need not have negligible total drag: its typical transverse drag scales as $\mu d/\log(d/b)$ for rod radius $b$. Thus the simple boxed result uses the intended negligibly dragging connector limit as well as $a_i/d\ll1$; the two printed size inequalities alone do not specify a uniform limit in which rod drag vanishes.
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