For the equation $5 x^{2}-a x+b=0$, the two roots are $\alpha, \beta$, and for $x^{2}-b x+c=0$, the two roots are $\gamma, \delta$ ($\alpha, \beta, \gamma, \delta$ are all distinct). Let $M=\{\alpha, \beta, \gamma, \delta\}, u, v \in M(u \neq v), S=\{x \mid x=u+v\}, P=\{x \mid x=u, v\}$. Also, $S=\{5, 7, 8, 9, 10, 12\}, P=\{6, 10, 14, 15, 21, 35\}$, find $a, b, c$.
Analysis and Solution We observe sets $P$ and $S$, and find that 10 is the only common element. By Vieta's formulas, we know $\alpha \beta = b = \gamma + \delta$. This gives us $a \beta = \gamma + \delta = b = 10$. Since the sum of the elements in set $S$ is 51, i.e., $3(\alpha + \beta + \gamma + \delta) = 51$, we get $\alpha + \beta = 7$. Therefore, $a = 7$. Also, the product of the elements in set $P$ is $210^3$, i.e., $(\alpha \beta \gamma \delta)^3 = 210^3$, so $c = 21$.