= Solution
The three structures are the radiative cascades
$$
\psi(2S)\to\gamma_l\chi_{cJ},\qquad \chi_{cJ}\to\gamma_h J/\psi,\qquad J=0,1,2.
$$
In the <quark model>, the <J psi meson> and $\psi(2S)$ are respectively $1\,{}^3S_1$ and $2\,{}^3S_1$ <charmonium> states. Thus $L=0$, $S=1$, $J=1$, and <meson parity and charge conjugation> gives $J^{PC}=1^{--}$. The intermediate <chi c mesons> are $1\,{}^3P_J$: $L=1$, $S=1$, and <addition of angular momentum> permits $J=0,1,2$. Consequently
$$
\boxed{\chi_{c0}:0^{++},\qquad \chi_{c1}:1^{++},\qquad \chi_{c2}:2^{++}.}
$$
These are <electric dipole transitions>, changing $L$ by one and preserving $S$ to leading order. A singlet $1\,{}^1P_1$ state has $J^{PC}=1^{+-}$ and is not another member of these spin-triplet cascades.
The primary <photon> <energy> is $(M_{2S}^2-M_{\chi_J}^2)/(2M_{2S})$, about $261$, $171$ and $128\ \mathrm{MeV}$ for $J=0,1,2$. The secondary <photon> remains more energetic even after its boost into the parent <rest frame>. Hence $m^2(J/\psi\gamma_h)=M_{\chi_J}^2$: with this quantity on the vertical axis, each <resonance> gives a horizontal band. A finite <decay width> and detector <measurement resolution> broaden each band; photon ordering removes the mirrored duplication of an unordered <Dalitz plot>.
The faint band beside the negative-slope diagonal boundary is $\psi(2S)\to J/\psi\,\pi^0$, followed by $\pi^0\to\gamma\gamma$. The <neutral pion> is predominantly $(u\bar u-d\bar d)/\sqrt2$, a ${}^1S_0$ <pseudoscalar meson> with $J^{PC}=0^{-+}$. This production transition violates <isospin>, accounting for its small rate. To locate the band, set $X=m^2(J/\psi\gamma_l)$ and $Y=m^2(J/\psi\gamma_h)$. <Four-momentum conservation> gives
$$
\boxed{X+Y=M_{2S}^2+M_{J/\psi}^2-m_{\gamma\gamma}^2.}
$$
For the <neutral pion>, $m_{\gamma\gamma}^2=m_{\pi^0}^2$ is small, putting the line just inside the $m_{\gamma\gamma}^2=0$ boundary. The allowed $J/\psi\,\eta$ channel, with the <eta meson> also $0^{-+}$, produces a parallel line further inside, at $m_{\gamma\gamma}^2=m_\eta^2$. The positive-slope edge $X=Y$ instead comes from ordering the two <photon energies>; proximity to that edge alone would not identify the two-photon state.
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