= Solution
In the two-body centre-of-mass frame, <spatial reflection> reverses the relative coordinate of a <quark>-<antiquark> pair. Its orbital state of angular momentum $L$ has <orbital parity> $(-1)^L$, as follows from $Y_{Lm}(-\widehat{\mathbf r})=(-1)^LY_{Lm}(\widehat{\mathbf r})$. Opposite constituent <intrinsic parities> multiply to minus one. Thus the <meson parity and charge conjugation> rules begin with
$$
\boxed{P=\eta_q\eta_{\bar q}(-1)^L=(-1)^{L+1}.}
$$
To obtain <charge conjugation>, consider a self-conjugate flavor state and write its <spin>-orbital wavefunction using ordered <creation operators> $b_\sigma^\dagger(\mathbf p)d_\tau^\dagger(-\mathbf p)$. <Charge conjugation> interchanges the <quark> and <antiquark> operators. Reordering them back to <quark>-first order produces a <fermionic sign> minus one. Interchanging the two <spin>-$1/2$ labels has eigenvalue $-1$ in the <spin singlet state> ($S=0$) and $+1$ in the <spin> triplet ($S=1$), namely $(-1)^{S+1}$. Reversing the relative momentum contributes $(-1)^L$. Therefore
$$
\boxed{C=-(-1)^{S+1}(-1)^L=(-1)^{L+S}.}
$$
Reciprocal phases for the two conjugate <creation operators> cancel, so the result is independent of that convention. This applies to a state that is mapped to itself by <charge conjugation>, not to every electrically neutral flavor state. For example a neutral <meson> with unequal <quark> and <antiquark> flavors can be exchanged with a different antimeson. Charged states have no individual $C$ eigenvalue.
Restrict flavor to the $u,d$ <isospin> doublet. Its product with the conjugate doublet decomposes as $\mathbf2\otimes\overline{\mathbf2}=\mathbf3\oplus\mathbf1$. Convenient normalized flavor states, with irrelevant overall phase choices, are
$$
|1,+1\rangle=u\bar d,\quad |1,0\rangle=\frac{u\bar u-d\bar d}{\sqrt2},\quad |1,-1\rangle=d\bar u,\quad
|0,0\rangle=\frac{u\bar u+d\bar d}{\sqrt2}.
$$
Orbital excitation normally costs energy, so the simplest low-lying <quark model> states have $L=0$. Coupling the two constituent spins gives $S=0$ or $S=1$, and then $J=S$. The isotriplet with $S=0$ is the <pion> triplet, while the isotriplet with $S=1$ is the <rho meson> triplet. The isosinglet with $S=1$ is the nonstrange component of the <omega meson>. Their quantum numbers are
$$
\boxed{\pi^0:\ I=1,\ J^{PC}=0^{-+};\qquad
\rho^0:\ I=1,\ J^{PC}=1^{--};\qquad
\omega:\ I=0,\ J^{PC}=1^{--}.}
$$
The charged <pions> have $J^P=0^-$ and the charged rho <mesons> have $J^P=1^-$; <charge conjugation> exchanges the positive and negative members. The same <tensor> product also allows an $S=0$, $I=0$ pseudoscalar state, so symmetry alone does not assert that only the listed <mesons> exist or predict their masses. The physical isosinglet spectrum involves further dynamics and, when strange flavor is included, mixing.
For a vector-<meson> decay to two spinless <pions>, conservation of <angular momentum> forces their relative orbital value to be $L=1$. Their combined <parity> is $(-1)^2(-1)^L=-1$, and the neutral charge combination has $C=(-1)^L=-1$, matching both $\rho^0$ and $\omega$. Thus $C$ and $P$ alone do not distinguish these two decays. The decisive restriction is <isospin> together with <Bose statistics>.
Two isotriplet <pions> have $\mathbf3\otimes\mathbf3=\mathbf1\oplus\mathbf3\oplus\mathbf5$, or total $I=0,1,2$. Under exchange the <isospin>-$I$ state has sign $(-1)^{2-I}$: the $I=2$ highest state and its lowered partners are symmetric, the $I=1$ highest combination $(|+0\rangle-|0+\rangle)/\sqrt2$ is antisymmetric, and the singlet $(|+-\rangle+|-+\rangle-|00\rangle)/\sqrt3$ is symmetric. The $L=1$ orbital state is antisymmetric. Since <pions> are <spin>-zero <bosons>, their total state must be symmetric, requiring the antisymmetric <isospin> sector $I=1$. This allows $\rho^0\to\pi^+\pi^-$ but excludes an <isospin>-conserving $\omega\to\pi^+\pi^-$ amplitude.
There is no analogous obstruction for three <pions>. In a Cartesian <isospin> basis $a,b,c$, their totally antisymmetric combination $\epsilon_{abc}|\pi^a\pi^b\pi^c\rangle$ is an $I=0$ state. It can multiply a totally antisymmetric momentum amplitude proportional in the rest frame to
$$
\boldsymbol\epsilon_\omega\cdot(\mathbf p_1\times\mathbf p_2).
$$
Momentum conservation $\mathbf p_1+\mathbf p_2+\mathbf p_3=0$ makes this amplitude change sign under every exchange of two <pion> labels. The product of the flavor and momentum factors therefore obeys <Bose statistics>. The cross product supplies $J=1$ and an even <orbital parity>; multiplying by the three negative <pion> <intrinsic parities> gives total $P=-1$. In the charged component, <charge conjugation> exchanges $\pi^+$ and $\pi^-$ and changes the momentum factor's sign, giving $C=-1$. Thus \b[$\omega\to\pi^+\pi^-\pi^0$ is compatible with $I=0$ and $J^{PC}=1^{--}$].
The same distinction is encoded in <G parity>: $G=C(-1)^I$ for the neutral member gives $G_\pi=-1$, $G_\rho=+1$, $G_\omega=-1$. An $n$-<pion> state has $G=(-1)^n$, so the <G-parity selection rule for pion multiplicities> allows two <pions> for the rho and three for the omega. The printed prohibition of the omega two-<pion> mode is an exact-<isospin> strong-interaction selection rule. In the physical theory, <isospin>-breaking effects allow a small two-<pion> mode; it is not an absolute prohibition. The scope of C <parity> is also stated in the https://pdg.lbl.gov/2024/reviews/rpp2024-rev-quark-model.pdf[Particle Data Group quark-model review], and the physical two-<pion> mode is recorded in its https://pdg.lbl.gov/2024/listings/rpp2024-list-omega-782.pdf[omega listing].
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