If the points $A\left(3,4\right),B\left(7,12\right)$ and $P\left(x,x\right)$ are such that ${\left(PA\right)}^2>{\left(PB\right)}^2>{\left(AB\right)}^2$ then the integral value of $x$ can be

If we simplify it, we get $x>7\to \left(1\right)$

${\left(PB\right)}^2>A{B}^2\Rightarrow {\left(x-7\right)}^2+{\left(x-12\right)}^2>16+64\Rightarrow 2x^2-38x+113>0$

Let $2x^2-38x+113=0\Rightarrow x=\frac{38\pm \sqrt{{38}^2-8\left(113\right)}}{4}=\frac{19\pm \sqrt{135}}{2}$

$=\frac{19\pm 11.6}{2}=\frac{30.6}{2}\left(or\right)\frac{7.4}{2}=15.3\left(or\right)3.7$

$\therefore \text{'}x\text{'}$  not lies between 15.3 and 3.7$\to \left(2\right)$

${\left(PA\right)}^2=A{B}^2\Rightarrow {\left(x-3\right)}^2+{\left(x-4\right)}^2>80\Rightarrow 2x^2-14x-55=0$

$\therefore \text{'}x\text{'}$  not lies between $-2.8\text{and}9.8\to \left(3\right)$

From (1), (2), (3)