Given,
$\begin{aligned} & \frac{2 r_2 r_3}{r_2-r_1}=r_3-r_1 \\ & \Rightarrow 2 r_2 r_3=\left(r_2-r_1\right)\left(r_3-r_1\right) \\ & \Rightarrow 2 \cdot \frac{\Delta}{(s-b)} \cdot \frac{\Delta}{(s-c)}=\left(\frac{\Delta}{s-b}-\frac{\Delta}{s-a}\right)\left(\frac{\Delta}{s-c}-\frac{\Delta}{s-a}\right) \\ & \Rightarrow \frac{2 \Delta^2}{(s-b)(s-c)}=\Delta^2\left\{\frac{s-a-s+b}{(s-b)(s-a)}\right\}\left\{\frac{s-a-s+c}{(s-c)(s-a)}\right\} \\ & \Rightarrow \frac{2}{(s-b)(s-c)}=\frac{(b-a)}{(s-b)(s-a)} \times \frac{(c-a)}{(s-c)(s-a)} \\ & \Rightarrow \quad 2(s-a)^2=(b-a)(c-a) \\ & \Rightarrow \frac{2(b+c-a)^2}{4}=(b-a)(c-a) \\ & \Rightarrow b^2+c^2+a^2+2 b c-2 c a-2 a b=2 \\ & \Rightarrow b^2+c^2+a^2+2 b c-2 c a-2 a b \\ & \Rightarrow b^2+c^2+a^2=2 a^2 \Rightarrow a^2=b^2+c^2 \\ & \text { Now } \frac{r_1\left(r_2+r_3\right)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}}\end{aligned}$
$\begin{aligned} & =\frac{\frac{\Delta}{s-a} \times \Delta \cdot\left\{\frac{1}{s-b}+\frac{1}{s-c}\right\}}{s} \\ & =\frac{\Delta^2(2 s-b-c)}{s(s-a)(s-a)(s-c)} \\ & =\frac{\Delta^2(a+b+c-b-c)}{\Delta^2}=a=2 R\end{aligned}$