400 followers problem!

$\text{Let P , Q , R be the reflections of H in BC , AC , AB respectively.}$

$m\angle BAC = a \quad, \quad m\angle ABC = b \quad , \quad m\angle ACB = c \\ a^| = 90-a \quad,\quad b^| = 90-b\quad,\quad c^| = 90-c$

By angle chasing we can clearly get that quadrilaterals ABPC , ABCQ , ARBC are cyclic since their opposite angles are supplementary.

$\implies R(ABC)=R(BPC)=R(AQC)=R(ARB)$

Since $\Delta BHC \cong \Delta BPC \quad\Delta AHC\cong \Delta AQC\quad\Delta AHB \cong \Delta ARB \\ \implies \boxed{R(ABC)=R(HAB)=R(HBC)=R(HAC)}$

Area of $\Delta ABC = 84$ (Using Heron's formula).Also ,

$\Delta ABC = \dfrac{\text{product of all sides}}{4R} \Rightarrow 4R = \dfrac{2730}{84} = 32.5$ where 'R' is circumradius of $\Delta ABC$

Area of $\Delta ABC = r_0.s \Rightarrow r_0 = \dfrac{84}{21} = 4$ where 's' is the semiperimeter of the triangle.

We also have the relation $r_1+r_2+r_3=r_0+4R$

$\displaystyle \sum_{i=0}^3 = 4R +2r_0= 32.5 + 8 = 40.5$

So , $\large R(ABC)+R(HAB)+R(HBC)+R(HAC)+\displaystyle \sum_{i=0}^3 r_i$

$\Rightarrow 4R + 4R + 2r_0 = 32.5+40.5 = \huge\boxed{73}$

@Azhaghu Roopesh M

In $\Delta ABC$ , $R(\Delta ABC)=\frac {14}{SinA}=\frac {15}{SinB}=\frac {13}{SinC}$

In $\Delta HBC$ , $R(\Delta HBC)=\frac {14}{Sin(B+C)}$

But we know $A+B+C=\pi$

Hence

$R(\Delta HBC)=\frac {14}{Sin(\pi-A)}=\frac {14}{SinA}$ = $R(\Delta ABC)$

Similarily ,

$R(\Delta ABC)$ = $R(\Delta HAC)$ = $R(\Delta HBA)$ = $R(\Delta HBC)$

$ar(\Delta ABC) =84 \Rightarrow R(\Delta ABC)=\frac {abc}{4ar(\Delta ABC)}$

$r_0=\frac{ar(\Delta ABC)}{s}$

$r_1+r_2+r_3=r_0+4R$

Hence we have to find $8R+2r$ = $\frac{8\times 13\times 14\times 15}{4\times 84}$ + $\frac{2\times 84}{21} = \boxed{73}$

Note: s is semiperimeter.

did by coordinates