= Solution
For operations ordered as $(E,C_2(z),\sigma_{xz},\sigma_{yz})$, the characters of the twelve Cartesian displacements are
$$
\chi_{3N}=(12,-2,4,2).
$$
Only atoms left fixed contribute: all four for $E$ and $\sigma_{xz}$, but only C and O for $C_2$ and $\sigma_{yz}$. Multiplying their counts by the Cartesian traces $3,-1,1,1$ gives these numbers. Subtract
$$
\chi_{\rm trans}=(3,-1,1,1),\qquad \chi_{\rm rot}=(3,-1,-1,-1)
$$
to obtain $\chi_{\rm vib}=(6,0,4,2)$. The <character table> reduction gives
$$
\boxed{\Gamma_{\rm vib}=3A_1+2B_1+B_2.}
$$
The dimensions sum to $6=3N-6$, as required. The $B_2$ mode is the out-of-plane wag with this axis convention.
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