= Solution
Choose time $t_c$ when the passing <sphere> centres are at the same height, and let $S$ reflect positions in that horizontal plane. For each instantaneous configuration, <Stokes flow> determines particle <velocities> uniquely and linearly from the fixed external <forces>. Time reversal reverses all <velocities> and the dense <sphere>'s applied <force>; reflecting in the horizontal plane reverses that <force> again. The neutral <sphere> remains force-free under both operations. Thus the reflected time-reversed motion obeys the same original forcing.
At $t_c$, reflection fixes both particle centres. Since the <spheres> have no orientational shape variables and there is no contact, uniqueness of the particle-trajectory initial-value problem makes the reflected backward motion coincide with the forward motion:
$$
\boxed{\mathbf r_j(t_c+\tau)=S\mathbf r_j(t_c-\tau),\qquad j=1,2.}
$$
This <reflection symmetry of a two-sphere passing trajectory> fixes the transverse coordinates and reverses the relative axial separation. Consequently, at matching large separations above and below, \b[both <spheres> return to their original distances from the centreline]. Their intermediate lateral excursions are undone. This uses <kinematic reversibility of Stokes flow>, geometric reflection symmetry and the noncontact result, not a dissipation-minimizing trajectory. It assumes the <spheres> pass one another; a perfectly collinear approach would not supply the side-by-side symmetry event.
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