$\dfrac{1}{r}=\dfrac{1}{r_A}+\dfrac{1}{r_B}+\dfrac{1}{r_C} \Rightarrow r=3$

$A(ABC)(\Delta)=\sqrt{r.r_A.r_B.r_C} \Rightarrow \Delta = 60$

$\dfrac{\Delta}{r_A}=(s-a) \\\dfrac{\Delta}{r_B}=(s-b) \\\dfrac{\Delta}{r_C}=(s-c)$

Adding all 3 gives :

$12+5+3=3s-2s \\ s=20 \\ \Phi=40 \\ \boxed{10\Phi=400}$

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From the properties of triangle, we know that

$r_1=\frac{\delta}{(s-a)}$ , $r_1=\frac{\delta}{(s-a)}$ and $r_1=\frac{\delta}{(s-a)}$

$s$ = semi-perimeter, $\delta$ = area and $a, b, c$ = sides of triangle

Now to obtain a relation between here we can do some rearrangements in above equation. $r_1\cdot r_2+r_2\cdot r_3+r_3\cdot r_1$ $=\frac{\delta ^2}{(s-a)\cdot(s-b)}+\frac{\delta ^2}{(s-b)\cdot(s-c)}+\frac{\delta ^2}{(s-a)\cdot(s-c)}$ $= \frac{\delta ^2}{(s-a)\cdot(s-b)\cdot(s-c)}\{{(s-a)+(s-b)+(s-c)}\}$ $= \frac{s\cdot\delta ^2}{s\cdot(s-a)\cdot(s-b)\cdot(s-c)}\{{s}\}$

So $r_1\cdot r_2+r_2\cdot r_3+r_3\cdot r_1 = s^2$

Hence $\Phi = 2\cdot s =2 \sqrt{r_1\cdot r_2+r_2\cdot r_3+r_3\cdot r_1 }$

Putting the values we get, $\Phi=40 \implies \boxed{10\Phi=400}$

SAME analysis