= Solution
Use the stated local axes: $z$ is radial along O–Ru, $x$ is tangential within the Ru–O plane, and $y$ is perpendicular to that plane. The <oxygen>'s three $sp^2$ hybrids make the in-plane sigma framework. Its remaining \b[$p_y$ orbital] is perpendicular to the plane and can overlap with \b[one $d_{yz}$ orbital on each Ru].
Choose the local orbital phases so their O-facing lobes agree. Of the three metal orbitals, only
$$
D=\frac{d_{yz}^{(1)}+d_{yz}^{(2)}+d_{yz}^{(3)}}{\sqrt3}
$$
has the appropriate combination to interact with <oxygen> $p_y$. The other two independent combinations have coefficients summing to zero and are nonbonding with respect to the central <oxygen>. Mixing $D$ with $p_y$ gives one lower pi-bonding orbital and one upper pi-antibonding orbital. Each Ru also retains its $d_{xy}$ and $d_{xz}$ orbitals as nonbonding with respect to this central-oxo pi interaction. Thus the complete low-energy diagram contains \b[one bonding level, eight metal-nonbonding levels, and one antibonding level]. The eight consist of two $d_{yz}$ combinations, three $d_{xy}$ and three $d_{xz}$ orbitals. They need not be exactly degenerate once other ligands are considered.
For the neutral formula as printed, oxide contributes $-2$ and six acetates contribute $-6$. The Ru oxidation numbers therefore sum to $+8$, giving sixteen metal d <electrons> overall, formally two Ru(III) and one Ru(II), possibly delocalized. Including the <oxygen> pi pair gives eighteen <electrons>, filling the <bonding orbital> and all eight <nonbonding orbitals>, leaving pi\* empty. If the often encountered singly charged cation were intended instead, there would be seventeen <electrons> and one singly occupied nonbonding level. This charge distinction changes occupancy, not the orbital connectivity.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1-oxo-mos.png]
{title=Qualitative central-oxo pi orbitals for the neutral triruthenium cluster}
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