An algebra problem by Dev Sharma

$Consider\quad \triangle ABC,with\quad AB=3,AC=5,BC=4.obviously,\quad it\quad is\quad a\quad right-angled\quad triangle.\\ O\quad is\quad a\quad piont\quad inside\quad the\quad such\quad that\quad \angle \quad AOB,\angle AOC,\angle BOC=120°\\ let\quad AO=a,OB=b,OC=c,by\quad cosine\quad law.\\ { a }^{ 2 }+{ b }^{ 2 }-2abcos120°={ 3 }^{ 2 }\rightarrow { a }^{ 2 }+{ b }^{ 2 }+ab=9\\ { b }^{ 2 }+{ c }^{ 2 }-2bccos120°={ 4 }^{ 2 }{ \rightarrow b }^{ 2 }+{ c }^{ 2 }+bc=16\\ { a }^{ 2 }+{ c }^{ 2 }-2accos120°={ 5 }^{ 2 }\rightarrow { a }^{ 2 }+{ c }^{ 2 }+ac=25,which\quad is\quad a\quad triangle\quad satisfies\quad the\quad equations\quad \\ By\quad consider\quad it's\quad area,\quad \frac { absin120° }{ 2 } +\frac { bcsin120° }{ 2 } +\frac { acsin120° }{ 2 } =\frac { 3\times 4 }{ 2 } \\ (ab+bc+ac)\frac { sin120° }{ 2 } =6\rightarrow (ab+bc+ac)=8\sqrt { 3 }$