In a  $\Delta ABC$ , with usual notation, if  $r=r_1-r_2-r_3$ , then  $2R=$

In a  $\Delta ABC$ , it is given that,
 $$\begin{gathered} r=r_1-r_2-r_3 \\ \Rightarrow \frac{\Delta }{s}=\frac{\Delta }{s-a}-\frac{\Delta }{s-b}-\frac{\Delta }{s-c} \\ \Rightarrow \frac{1}{s-b}+\frac{1}{s-c}=\frac{1}{s-a}-\frac{1}{s} \\ \Rightarrow \frac{2s-b-c}{(s-b)(s-c)}=\frac{a}{s(s-a)} \\ \Rightarrow s(s-a)=s^2-(b+c)s+bc \\ \Rightarrow s^2-sa=s^2-(b+c)s+bc \\ \Rightarrow (b+c-a)s=bc\Rightarrow {(b+c)}^2-a^2=2bc \\ \Rightarrow b^2+c^2+2bc-a^2=2bc \\ \therefore \Delta ABC is the right angled triangle with angle A={90}^0 , so 2R=a. \end{gathered}$$