Let $a,b,c$ be the length of three sides of a triangle satisfying the condition $\left(a^2+b^2\right)x^2-2b\left(a+c\right)$ $x+\left(b^2+c^2\right)=0$. If the set of all possible values of $x$ is in the interval $\left(\alpha ,\beta \right),$ then $12\left({\alpha }^2+{\beta }^2\right)$ is equal to _______.

Given,

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

$\Rightarrow a^2{x}^2-2abx+b^2+b^2{x}^2-2bcx+c^2=0$

$\Rightarrow {\left(ax-b\right)}^2+{\left(bx-c\right)}^2=0$

$\Rightarrow ax-b=0, bx-c=0$

$\Rightarrow ax=b, bx=c$

Now, we know that the sum of two sides is always greater than the third side of a triangle,

Now, taking $a+b>c$ we get,

$\Rightarrow a+ax>bx$

$\Rightarrow a+ax>a{x}^2$

$\Rightarrow x^2-x-1<0$

$\Rightarrow \frac{1-\sqrt{5}}{2}<x<\frac{1+\sqrt{5}}{2} ...\left(i\right)$

Similarly, for $b+c>a$ we get,

$\Rightarrow x^2+x-1>0$

$\Rightarrow x\in \left(-\infty ,\frac{-1-\sqrt{5}}{2}\right)\cup \left(\frac{-1+\sqrt{5}}{2},\infty \right) ....\left(ii\right)$

And for $c+a>b$ we get,

$\Rightarrow x^2-x+1>0$

$\Rightarrow x\in R ........\left(iii\right)$

So, from the equation $\left(i\right), \left(ii\right) \& \left(iii\right)$ we get,

$x\in \left(\frac{-1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}\right)$

$\Rightarrow \alpha =\frac{\sqrt{5}-1}{2}, \beta =\frac{\sqrt{5}+1}{2}$

Hence, $12\left({\alpha }^2+{\beta }^2\right)=12\left(\frac{{\left(\sqrt{5}-1\right)}^2+{\left(\sqrt{5}+1\right)}^2}{4}\right)=36$