For $0<c<b<a$, let $(a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0$ and $\alpha\neq1$ be one of its roots. Then, among the two statements: (I) If $\alpha\in(-1,0)$, then $b$ cannot be the geometric mean of $a$ and $c$. (II) If $\alpha\in(0,1)$, then $b$ may be the geometric mean of $a$ and $c$.
If a, b, c, p, q, r are non-zero real numbers, such that a < b < c and\[f(x) = (x - a)(x - b)(x - c) - p^2(x - a) - q^2(x - b) - r^2(x - c),\]then \(f(x) = 0\) must have