Solution:
Suppose $a + b + c \neq 0$ and let the common value be $\lambda$. Then
$$\lambda = \frac{x b + (1-x) c + x c + (1-x) a + x a + (1-x) b}{a + b + c} = 1$$
We get two equations:
$$-a + x b + (1-x) c = 0, \quad (1-x) a - b + x c = 0$$
(The other equation is a linear combination of these two.) Using these two equations, we get the relations
$$\frac{a}{1-x + x^{2}} = \frac{b}{x^{2} - x + 1} = \frac{c}{(1-x)^{2} + x}$$
Since $1-x + x^{2} \neq 0$, we get $a = b = c$.