Suppose that two of the numbers had the same leading coefficient, say $ax^2+bx+c=ax^2+cx+b$. This gives \[(b-c)(x-1)=0\iff x=1\] Suppose that two of the numbers had the same coefficient of $x$, say $ax^2+bx+c=cx^2+bx+a$. This gives \[(a-c)(x^2-1)=0\iff x=\pm 1\]However, if $x=-1$, our numbers are $a-b+c,a-c+b,b-c+a,b-a+c,c-a+b,c-b+a$, no $3$ of which are equal. Thus $x=1$. Suppose that two of the numbers had the same constant term, say $ax^2+bx+c=bx^2+ax+c$. This gives \[(a-b)(x^2-x)=0\iff x=0,1\]However, if $x=0$, our numbers are $a,b,a,c,b,c$, no $3$ of which are equal. Thus $x=1$. The last case is when all $3$ are in a cycle, or $ax^2+bx+c=bx^2+cx+a=cx^2+ax+b$. Then, we get \[(a-b)x^2+(b-c)x+(c-a)=0\]\[(b-c)x^2+(c-a)x+(a-b)=0\]The roots of the first are $1,\dfrac{c-a}{a-b}$ while the roots of the second are $1,\dfrac{a-b}{b-c}$. If \[\dfrac{c-a}{a-b}=\dfrac{a-b}{b-c}\]Then, we get that \[\dfrac{b-c}{a-b}=\dfrac{c-a}{b-c}\]so thus \[(a-b)^2=(b-c)(c-a),(b-c)^2=(a-b)(c-a)\]multiplying we get \[(c-a)^2=(a-b)(b-c)\]However, we know the only solution to $x^2=yz,y^2=xz,z^2=xy$ is $x=y=z$. Thus, we get the common solution is $1$.