What is constant in the simultaneous fit? This should defined by PDG, whose uncertainties should be small enough:
$$ \mathcal{B}(K_1\to K^*_0(1430)\pi)=(28\pm4)\%,\\\mathcal{B}(K_1\to K\omega)=(11.0\pm 2.0)\%, \\ \mathcal{B}(K_1\to K\rho)=(38\pm13)\% $$
Target:
$$ \mathcal{\alpha}=\frac{\mathcal{B}(K_1\to K^*\pi)}{\mathcal{B}(K_1\to K\rho)} $$
where $\mathcal{B}(K_1\to K^\pi)=\mathcal{B}(K_1\to K^(1430)\pi)+\mathcal{B}(K_1\to K^*(892)\pi)$.
Through ISO spin, the branching fraction of physical observable is:
$$ \mathcal{B}(K_1^+\to K^+\pi^-\pi^+)=\frac{1}{3}\times \mathcal{B}(K_1\to K\rho)+\frac{4}{9}\times \mathcal{B}(K_1\to K^(892)\pi)+\frac{4}{9}\times \mathcal{B}(K_1\to K^_0(1430)\pi)+\mathcal{B}(K_1\to KX^0) $$
$$ \mathcal{B}(K_1^0\to K^+\pi^-\pi^0)=\frac{2}{3}\times \mathcal{B}(K_1\to K\rho)+\frac{4}{9}\times \mathcal{B}(K_1\to K^(892)\pi)+\frac{4}{9}\times \mathcal{B}(K_1\to K^_0(1430)\pi) $$
where $\mathcal{B}(K_1\to KX^0)= \mathcal{B}(K_1\to K\omega)+\mathcal{B}(K_1\to Kf_0(1370))$.
If we ignore the coefficients, and define it as:
$$ \mathcal{B}_{3body}=\mathcal{B}(K_1\to K\rho)+\mathcal{B}(K_1\to K^*\pi)+\mathcal{B}(K_1\to KX^0) $$
We consider two cases:
Define a $\beta$:
$$ \beta^{-1} = 1 - \frac{\mathcal{B}(K_1^+\to K^+\pi^-\pi^+)}{\mathcal{B}(K^0_1\to K^+\pi^-\pi^0)}\Rightarrow \alpha=\frac{3}{4}(\beta-2) $$