Isoperimetric points and Tetrahedrons.

Refer to isoperimetric point .

The perimeter $P^{*} = 13 + b + c = 14 + b + d = 15 + c + d \implies b = d + 2 \implies c = d + 1$

Using Heron's formula $A_{APC} = \sqrt{42(d + 8)(d - 5)}, A_{APB} = \sqrt{48(d + 8)(d - 6)}, A_{BPC} = \sqrt{56(d + 8)(d - 7)}$

and

$A_{ABC} = 84 \implies 84 = \sqrt{d + 8}(\sqrt{42(d - 5)} + \sqrt{48(d - 6)} +\sqrt{56(d - 7)})$

Solving for $d$ we obtain: $d = \dfrac{80}{11} \implies \overline{AP} = \dfrac{102}{11}, \overline{CP} = \dfrac{91}{11}$ and $\overline{BP} = \dfrac{80}{11}$ .

Let $\overline{BD}$ be altitude in $\triangle{ABC}:$

$A_{ABC} = 84 = \dfrac{13}{2}(\overline{BD}) \implies \overline{BD} = \dfrac{168}{13} \implies \overline{AD} = \dfrac{70}{13}$ .

Let $A(0,0), \:\ B(\dfrac{70}{13},\dfrac{168}{13}),\:\ C(13,0)$ and $P(x_{0},y_{0})$ .

Using $\triangle{APC} \implies$

$x_{0}^2 + y_{0}^2 = \dfrac{10404}{121}$

$x_{0}^2 - 26x_{0} + 169 = \dfrac{8281}{121}$

$\implies x_{0} = \dfrac{1026}{143}$ and $y_{0} = \dfrac{840}{143}$

Point $P(\dfrac{1026}{143},\dfrac{840}{143})$ .

$D(\dfrac{1026}{143},0) \implies \overline{PD} = \dfrac{840}{143} \implies \overline{QD} = \dfrac{\sqrt{20449h^2 + 705600}}{143} \implies A_{1} = A_{\triangle{AQC}} = \dfrac{1}{22}\sqrt{20449h^2 + 705600}$

For $A_{\triangle{QAB}}:$

$m_{AB} = \dfrac{84}{35} \implies 455y - 1092x = 0$

$\overline{PE} \perp \overline{AB} \implies$ $m_{\overline{PE}} = \dfrac{-35}{84} \implies 12012y + 5005x = 106470 \implies x = \dfrac{450}{143}$ and $y = \dfrac{1080}{143}$

$E( \dfrac{450}{143},\dfrac{1080}{143}) \implies \overline{PE} = \dfrac{624}{143} \implies \overline{QE} = \dfrac{\sqrt{20449h^2 + 389376}}{143}$

$\implies A_{2} = A_{\triangle{QAB}} = \dfrac{7}{143}\sqrt{20449h^2 + 389376}$

For $A_{\triangle{QBC}}:$

$m_{\overline{BC}} = -\dfrac{56}{33} \implies 33y + 56x = 728$

$\overline{PF} \perp \overline{Bc} \implies m_{\overline{PF}} = 8008y - 4719x = 13182 \implies x = \dfrac{2902}{325}$ and $y = \dfrac{24696}{3575}$

$F(\dfrac{2902}{325},\dfrac{24696}{3575}) \implies \overline{PF} = \dfrac{7280}{3575} \implies \overline{QF} = \dfrac{\sqrt{12780625h^2 + 52998400}}{3575}$ $\implies A_{3} = A_{\triangle{QBC}} = \dfrac{3}{1430}\sqrt{12780625h^2 + 52998400}$ .

and,

$A_{4} = A_{\triangle{ABC}} = 84 \implies$ The volume of the tetrahedron $V = 28h = 224 \implies h = 8$

$\implies$

$A_{1} = \dfrac{\sqrt{2014336}}{22} \approx \boxed{64.512412}$

$A_{2} = \dfrac{7\sqrt{1698112}}{143} \approx \boxed{63.788908}$

$A_{3} = \dfrac{3\sqrt{870958400}}{1430} \approx \boxed{61.913296}$

and,

$A_{4} = \boxed{84}$

$\implies$ The total surface area $S = \boxed{274.214616}$ .