= Solution
\b[A <triangular prism>] has exactly the required nine metal–metal edges: six edges around two triangular faces and three edges joining corresponding vertices of those faces. It naturally offers two symmetry-distinct edge families, of sizes six and three.
With twelve $\mathrm{Cl^-}$ ligands, neutral $\mathrm{Re_6Cl_{12}}$ has total rhenium oxidation number $+12$, hence six formally Re(II), $d^5$ centres. After treating the Re–Cl interactions as ligand bonds, the available metal $d$ <electrons> are
$$
6(7-2)=30\ \mathrm{electrons}=15\ \mathrm{bonding\ pairs}.
$$
Two localized <bond order> assignments use those fifteen pairs:
$$
\begin{array}{c|cc|c}
&\text{six triangular edges}&\text{three prism edges}&\text{total bond order}\\
\mathrm{I}&1&3&6+9=15\\
\mathrm{II}&2&1&12+3=15
\end{array}
$$
In scheme I the three inter-triangle edges should be shorter; in scheme II the six <triangle> edges should be shorter. Both give each Re five <electrons>' worth of metal–metal bond-order contributions: two singles plus a triple, or two doubles plus a single. \b[The <electron> count permits both schemes]; identifying which family is shorter distinguishes them at this level. Actual cluster <molecular orbitals> are delocalized, so these integer <bond orders> summarize bonding rather than asserting isolated textbook multiple bonds on every edge.
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