Proving that if these quadratics are equal for some $\alpha$, then their coefficients are equal Let 
$$P_1(x) = ax^2 -bx - c \tag{1}$$
$$P_2(x) = bx^2 -cx -a \tag{2}$$
$$P_3(x) = cx^2 -ax - b \tag{3}$$
Suppose there exists a real $\alpha$ such that 
$$P_1(\alpha) = P_2(\alpha) = P_3(\alpha)$$
Prove 
$$a=b=c$$
Equating $P_1(\alpha)$ to $P_2(\alpha)$ 
$$\implies a\alpha^2 - b\alpha - c = b\alpha^2 - c\alpha - a$$
$$\implies (a-b)\alpha^2 + (c-b)\alpha + (a-c) = 0$$
Let 
$$Q_1(x) = (a-b)x^2 + (c-b)x + (a-c)$$
This implies, $\alpha$ is a root of $Q_1(x)$.
Similarly, equating $P_2(\alpha)$ to $P_3(\alpha)$ and $P_3(\alpha)$ to $P_1(\alpha)$, and rearranging we obtain quadratics $Q_2(x)$ and $Q_3(x)$ with common root $\alpha$:
$$Q_2(x) = (b-c)x^2 + (a-c)x + (b-a)$$
$$Q_3(x) = (c-a)x^2 + (b-a)x + (c-b)$$
$$Q_1(\alpha) = Q_2(\alpha) = Q_3(\alpha) = 0$$
We have to prove that this is not possible for non-constant quadratics $Q_1(x), Q_2(x), Q_3(x)$. 
EDIT:
I also noticed that for distinct $a, b, c \in \{1, 2, 3\}$:
$$Q_a(x) + Q_b(x) = -Q_c(x)$$

Denote
$$Q_1(x)=P_1(x)-P_2(x)=(a-b)x^2-(b-c)x-(c-a);$$
$$Q_2(x)=P_2(x)-P_3(x)=(b-c)x^2-(c-a)x-(a-b);$$
$$Q_3(x)=P_3(x)-P_1(x)=(c-a)x^2-(a-b)x-(b-c).$$
Then $\alpha$ is a real root of the equations $Q_i(x);$ so that $\Delta_{Q_i(x)}\geq 0 \ \ \forall i=1,2,3;$ where $\Delta_{f(x)}$ denoted the discriminant of a quadratic function $f$ in $x.$
Now, using this we have,
$$(b-c)^2+4(a-b)(c-a)\geq 0;$$
$$(c-a)^2+4(a-b)(b-c)\geq 0;$$
$$(a-b)^2+4(b-c)(c-a)\geq 0.$$
Summing these up and using the identity
$$\begin{aligned}(a-b)^2+(b-c)^2+(c-a)^2+2\left(\sum_\text{cyc}(a-b)(c-a)\right)\\=[(a-b)+(b-c)+(c-a)]^2=0;\end{aligned}$$
We obtain,
$$\begin{aligned}0&\leq 2\left(\sum_\text{cyc}(a-b)(c-a)\right)=2\left(\sum_\text{cyc}(ca+ab-bc-a^2)\right)\\&=-\left[(a-b)^2+(b-c)^2+(c-a)^2\right];\end{aligned}$$
Possible if and only if $(a-b)^2+(b-c)^2+(c-a)^2=0\iff a=b=c.$ We are done
