3 unknowns, 1 equation

$\begin{aligned} a\left(\frac{a^2}{b-a}+\frac{b^2}{c-b}+\frac{c^2}{2a-c}\right)&=\left(\left(b-a\right)+\left(c-b\right)+\left(2a-c\right)\right)\left(\frac{a^2}{b-a}+\frac{b^2}{c-b}+\frac{c^2}{2a-c}\right) \\ &\ge \left(\sqrt{\frac{a^2}{b-a}\left(b-a\right)}+\sqrt{\frac{b^2}{c-b}\left(c-b\right)}+\sqrt{\frac{c^2}{2a-c}\left(2a-c\right)}\right)^2 \quad (*) \\ &= (a+b+c)^2 \end{aligned}$ (*) By the Cauchy–Schwarz inequality

Since the question gives that equality holds:

$\frac{a^2}{\left(b-a\right)^2}=\frac{b^2}{\left(c-b\right)^2}=\frac{c^2}{\left(2a-c\right)^2}$

$\frac{b}{a}=\frac{c}{b}=\frac{2a}{c}$

$\frac{bc}{a^2}=\boxed{2}$