We have,
$R: r_1: r: 5: 12: 2$
$\therefore R=5 k, r_1=12 k, r=2 k$
If $\triangle A B C$ is right angle of $A$ and $a=10 k, b=6 k, c=8 k$
Then $R=5 k, r_1=12 k$ and $r=2 k$
Possible
$\therefore \quad a^2=b^2+c^2$
$\therefore \quad r+r_3+r_2-r_1$


$\frac{\Delta}{s}+\frac{\Delta}{s-c}+\frac{\Delta}{s-b}+\frac{\Delta}{s-a}$
$\left(\frac{\Delta}{s}-\frac{\Delta}{s-a}\right)+\left(\frac{\Delta}{s-c}+\frac{\Delta}{s-b}\right)$
$\Rightarrow \Delta\left(\frac{s-a-s}{s(s-a)}+\frac{s-b+s-c}{(s-c)(s-b)}\right)$
$\Rightarrow \Delta\left[\frac{-a}{s(s-a)}+\frac{2 s-b-c}{(s-c)(s-b)}\right]$
$\Rightarrow \Delta\left[\frac{-a}{s(s-a)}+\frac{a}{(s-c)(s-b)}\right]$
$\Rightarrow \quad \Delta a\left[\frac{-(s-c)(s-b)+s(s-a)}{s(s-a)(s-b)(s-c)}\right]$
$\Rightarrow \Delta a \frac{\left(s^2-a s-s^2+(b+c) s-b c\right)}{\Delta^2}$
$\Rightarrow \frac{a^2-\left(b^2+c^2\right)}{\Delta^2}=0=\cos A$
$\therefore \cos A$