If $a,b,c$ are positive rational numbers such that $a>b>c$ and the quadratic equation $\left(a+b-2c\right)x^2+\left(b+c-2a\right)x+\left(c+a-2b\right)=0$ has a root in the interval $\left(-1,0\right)$ then

We have $a>b>c\to \left(i\right)$

$\left(a+b-2c\right)x^2+\left(b+c-2a\right)x+\left(c+a-2b\right)=0\to \left(ii\right)$

$\because \left(ii\right)$ has a root in the interval $\left(-1,0\right)$ we have $f\left(-1\right).f\left(0\right)<0$

$\left(2a-b-c\right)\left(c+a-2b\right)<0\to \left(iii\right)$

From (i) $a>b\&a>c$

$\Rightarrow a-b>0\Rightarrow a-c>0$

From (iii)  & (iv)  $c+a-2b<0$

$\Rightarrow c+a<2b$

$\because$ Sum of coefficients in (ii) is zero

$\because$  roots are 1 & $\frac{c+a-2b}{a+b-2c}$

Determinant of $a{x}^2+2bx+c=0$ is $4b^2-4ac=D$

$c+a<2b$

$\begin{array}{l}\Rightarrow 4b^2>{\left(c+a\right)}^2\\ 4b^2-4ac>{\left(c+a\right)}^2-4ac\\ D>{\left(c-a\right)}^2\\ D>0\end{array}$

Also each of $a,b,c$ is positive

$\therefore$$a{x}^2+2bx+c=0$ has real negative roots.

Similary $c{x}^2+2ax+b=0$ has both negative real roots.