= Solution
\b[Yes: let the order be voter 2, then voter 3, then voter 1.] When voter 1 moves last, it makes $A$ win whenever that is possible; otherwise it prefers $B$ to $C$. For the first two votes in this new order, its optimal continuation winners are
$$
\begin{array}{c|ccc}
&A&B&C\\\hline
A&A&A&A\\
B&A&B&A\\
C&A&A&C
\end{array}.
$$
Rows now record voter 2's initial choice and columns voter 3's choice. In a three-way split, the specified tie-break produces $A$, which explains, for example, the off-diagonal $B,C$ entry.
Voter 3 optimizes its ranking $B\succ C\succ A$ in each row, giving the continuation winners $A,B,C$. Voter 2 then chooses $C$, its favourite of the three. The winner is therefore \b[$C$ in every <subgame perfect equilibrium>], again by <unique outcome under strict outcome preferences>. In particular, if voter 2 starts with $C$, voter 3 must also vote $C$: either other vote lets the last voter force $A$.
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