System of Equations

$Solution\quad by\quad the\quad organisers\quad of\quad SMO\quad (Junior)\quad 2011:\\ Consider\quad \triangle ABC\quad where\quad AB=5,\quad BC=7\quad and\quad AC=8.\\ Let\quad P\quad be\quad the\quad point\quad in\quad \triangle ABC\quad such\quad that\quad \angle APB=\angle BPC=\angle CPA=120°.\\ Then,\quad by\quad Cosine\quad Rule,\quad we\quad can\quad let\quad AP=a,\quad BP=b\quad and\quad CP=c.\\ The\quad area\quad of\quad the\quad triangle\quad is\quad \sqrt { (10)(5)(3)(2) } =10\sqrt { 3 } .\\ Hence,\quad by\quad sine\quad formula\quad for\quad area\quad of\quad a\quad triangle,\quad \frac { \sqrt { 3 } }{ 4 } (ab+bc+ca)=10\sqrt { 3 } .\\ This\quad implies\quad that\quad ab+bc+ca=40.\\ Summing\quad up\quad the\quad 3\quad given\quad equations,\quad we\quad get\quad 2{ a }^{ 2 }+2{ b }^{ 2 }+2{ c }^{ 2 }+ab+bc+ca=138.\\ Hence\quad 2{ (a+b+c) }^{ 2 }=138+3(40)=258.\quad So\quad { (a+b+c) }^{ 2 }=\boxed { 129 } .\\$