Let
$$\begin{align*} ax^2 + bx + c &= x_1, & ax^2 + cx + b &= x_2 \\ bx^2 + cx + a &= x_3, & bx^2 + ax + c &= x_4 \\ cx^2 + ax + b &= x_5, & cx^2 + bx + a &= x_6 \end{align*}$$
Then we get
$$\begin{align*} (b-c)(x-1) &= x_1 - x_2, \quad (c-a)(x-1) = x_3 - x_4, \quad (a-b)(x-1) = x_5 - x_6 \\ (b-c)(x^2-1) &= x_4 - x_5, \quad (c-a)(x^2-1) = x_6 - x_1, \quad (a-b)(x^2-1) = x_2 - x_3 \\ (b-c)(x^2-x) &= x_3 - x_6, \quad (c-a)(x^2-x) = x_5 - x_2, \quad (a-b)(x^2-x) = x_1 - x_4. \end{align*}$$
For $x = 0$, the set $\{x_1, x_2, x_3, x_4, x_5, x_6\} = \{c, b, a, c, b, a\}$ can not include three identical elements. Similarly, for $x = -1$, the set $\{x_1, x_2, x_3, x_4, x_5, x_6\} = \{a + c - b, a - c + b, b - c + a, b - a + c, c - a + b, c - b + a\}$ can not include three identical elements since if two of $a + b - c$, $b + c - a$, $c + a - b$ are equal, then two of $a$, $b$, $c$ are equal which is not possible. Consider the case $x \neq -1, 0, 1$. Since $a$, $b$, $c$ are distinct, using the equations above, we get $x_u \neq x_v$ for any odd $u$ and even $v$. Therefore the sets $S_1 = \{x_1, x_3, x_5\}$ and $S_2 = \{x_2, x_4, x_6\}$ should be disjoint. Therefore if there are three identical elements in $S_1 \cup S_2$, then either $x_1 = x_3 = x_5$ or $x_2 = x_4 = x_6$. W.L.O.G. assume that $x_1 = x_3 = x_5$. In this case, we have
$$ax^2 + bx + c = bx^2 + cx + a = cx^2 + ax + b.$$

$x = 1$ is a common root of the three 2-nd order polynomials given above, and hence we obtain that
$$(x-1)(x-t_1) = (x-1)(x-t_2) = 0$$
where $t_1 = \frac{c-a}{a-b}$, $t_2 = \frac{a-b}{b-c}$. There must be another common root different from $1$, and hence
$$\frac{c-a}{a-b} = \frac{a-b}{b-c}$$
which is equivalent to
$$(a-b)^2 + (b-c)^2 + (c-a)^2 = 0$$
which is not possible since $a$, $b$, $c$ are distinct. Therefore we get $x = 1$.