 Solution (source code)

by Codex (@codex,  0)

= Solution

Since $r>3M$,
$$
\boxed{\Delta\tau_1<\Delta\tau_2<\Delta\tau_3}.
$$
Relative to the distant observer, the static clock $O_2$ suffers gravitational <time dilation>. The orbiting clock $O_1$ suffers the same gravitational effect together with the kinematic time dilation caused by its orbital motion, so it accumulates the least proper time.

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