= Solution
Direct formation in $e^+e^-$ through one virtual <photon> selects the <vector meson> quantum numbers $J^{PC}=1^{--}$. The nonzero $K_S^0K_L^0$ mode and absence of $K_S^0K_S^0$ are consistent with this assignment: the two spin-zero <kaons> require odd <orbital angular momentum> for negative <parity>, whereas identical $K_S^0$ <bosons> cannot have an odd relative wave. This argument neglects the tiny <CP violation> in neutral <kaons>.
Assuming the $\rho\pi$ decays conserve <isospin>, their equal rates indicate $I=0$. The <Clebsch-Gordan coefficients> give
$$
|0,0\rangle=\frac{|\rho^+\pi^-\rangle-|\rho^0\pi^0\rangle+|\rho^-\pi^+\rangle}{\sqrt3}.
$$
An $I=1,I_3=0$ state lacks $\rho^0\pi^0$, while $I=2,I_3=0$ has squared coefficients $1:4:1$ rather than $1:1:1$. The branching fractions alone determine neither $J$ nor these relative phases; the single-<photon> formation and approximate <strong interaction> <isospin> conservation are essential additional inputs.
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