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 $a, b, c \in \{1, 2, 3\}$:
$$Q_a(x) + Q_b(x) = -Q_c(x)$$
 A: The resultant of $P_1(x) - P_2(x)$ and $P_1(x) - P_3(x)$ with respect to $x$ is
$2(a^2+b^2 + c^2-ab-ac-bc)^2$, so this must be $0$ if there is $\alpha$ such that
$P_1(\alpha) = P_2(\alpha) = P_3(\alpha)$.  Now the quadratic form $a^2+b^2 + c^2-ab-ac-bc$ is 
positive semidefinite: in fact it is $((a-b)^2 + (a-c)^2 + (b-c)^2)/2$, so it can only 
be $0$ (for real $a,b,c$) if $a=b=c$.
A: 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
A: If we assume that $ P_1(\alpha)=P_2(\alpha)=P_3(\alpha)=p $ :
\begin{align*}
p &= a\alpha^2 -b\alpha -c \qquad &&p= a\alpha^2 -b\alpha -c  \\ 
p &= b\alpha^2 -c\alpha -a \qquad &&p= -a +b\alpha^2 -c\alpha \\
p &= c\alpha^2 -a\alpha -b \qquad &&p= -a\alpha -b-c\alpha^2\\
\end{align*}
then we get the Cramer system : 
\begin{align*}
& p\begin{bmatrix}
       1 \\[0.3em]
       1 \\[0.3em]
       1 
     \end{bmatrix} = \begin{bmatrix}
       \alpha^2 & -\alpha & -1           \\[0.3em]
       -1 & \alpha^2           & -\alpha \\[0.3em]
       -\alpha           & -1  & \alpha^2
     \end{bmatrix}\begin{bmatrix}
            a \\[0.3em]
            b \\[0.3em]
            c 
          \end{bmatrix}\\
          &pI_{3}=A_{3,3}X_{3}
\end{align*}
if $ A_{3,3} $ is invertible then
\begin{equation*}
X_{3}=pA_{3,3}^{-1}I_{3}
\end{equation*}
Then the values of $ a $, $ b $ and $ c $ can be found as follows:
\begin{equation*}
a = p\frac { \begin{vmatrix} {\color{red}1} & -\alpha & -1   \\ {\color{red}1} & \alpha^2 & -\alpha \\ {\color{red}1} & -1  & \alpha^2 \end{vmatrix} } { \mid A_{3,3} \mid}, \quad b = p\frac { \begin{vmatrix} \alpha^2 & {\color{red}1} & -1  \\ -1 & {\color{red}1}           & -\alpha \\ -\alpha           & {\color{red}1}  & \alpha^2 \end{vmatrix} } {\mid A_{3,3} \mid},\text{ and }c = p\frac { \begin{vmatrix} \alpha^2 & -\alpha & {\color{red}1} \\-1 & \alpha^2           & {\color{red}1} \\-\alpha & -1  & {\color{red}1}\end{vmatrix} } {\mid A_{3,3} \mid}
\end{equation*}
with $ \det(A_{3,3})=\mid A_{3,3} \mid=\alpha^{6}-3\alpha^{3}+\alpha^{2}-1 $ then we gwt the answer :
\begin{equation*}
a=b=c=p\left[ \dfrac{\alpha^{4}+\alpha^{3}+2\alpha^{2}-\alpha+1}{\alpha^{6}-3\alpha^{3}+\alpha^{2}-1 }\right] 
\end{equation*}
