In ∆ A B C , suppose the radius of the circle opposite to angle A is denoted by r 1 , similarly r 2…

Given, in $∆ABC, r_1=2,r_2=3,r_3=6$

To find: $a, b, c=?$

$r_1=\frac{∆}{s-a},r_2=\frac{∆}{s-b},r_3=\frac{∆}{s-c}$

$s-a=\frac{∆}{2},s-b=\frac{∆}{3},s-c=\frac{∆}{6}$

Adding all,

$3s-(a+b+c)=\frac{∆}{2}+\frac{∆}{3}+\frac{∆}{6}$

$\Rightarrow 3s-2s=\frac{6∆}{6}$

$\Rightarrow s=∆ ......\left(i\right)$

As we know

$∆=\sqrt{s(s-a)(s-b)(s-c)}$

$\Rightarrow ∆=\sqrt{∆\frac{∆}{2}\frac{∆}{3}\frac{∆}{6}}$

$\Rightarrow ∆=\frac{{∆}^2}{6}$

$\Rightarrow ∆=6$

$\Rightarrow ∆=s=6$

$s-a=\frac{∆}{2},s-b=\frac{∆}{3},s-c=\frac{∆}{6}$

$6-a=\frac{6}{2},6-b=\frac{6}{3},6-c=\frac{6}{6}$

$6-a=3,6-b=2,6-c=1$

$a=3,b=4,c=5$