Just do it bro

Without losing the generally,we may assume:c=min{a,b,c}and $\rightarrow 0\leq c< a,b$ We have: $Q\geq[(a+b)^2+a^2+b^2][\frac{1}{(a-b)^2}+\frac{1}{a^2}+\frac{1}{b^2}]=2\frac{a^2+ab+b^2}{(a-b)^2}+2(\frac{a}{b}+\frac{b}{a})^2+2(\frac{a}{b}+\frac{b}{a})$ Let $t=\frac{a}{b}+\frac{b}{a}$ , $t> 2$ so $Q\geq f(t)=2(t^2+t+\frac{t+1}{t-2})$ Let considering function $f(t)=2(t^2+t+\frac{t+1}{t-2})$ in the range $(2;+\infty )$ we have: $f'(t)=2[2t+1-\frac{3}{(t-2)^2}]$ $f'(t)=0\Leftrightarrow (2t+1)(t-2)^3-3=0$ $\Leftrightarrow 2t^3-7t^2+4t+1=0\Leftrightarrow t=\frac{5+\sqrt{33}}{4}(t> 2)$ Because of $t=\frac{5+\sqrt{33}}{4}$ so f'(t) reversal from negative to positive so $f(t)_{min}$ only when $t=\frac{5+\sqrt{33}}{4}$ So $f(t)\geq f(\frac{5+\sqrt{33}}{4})=\frac{59+11\sqrt{3}}{4}$ $f(t)_{min}=\frac{59+11\sqrt{33}}{4}$ .The equality holds when $\frac{a}{b}+\frac{b}{a}=\frac{5+\sqrt{33}}{4},c=0$ .Note that a,b,c maybe permutation. So $Q=\frac{59+11\sqrt{33}}{4}=30.547547278$ ;and $\left \lfloor 1000Q \right \rfloor=\left \lfloor 30547.547278 \right \rfloor=30547$