In a triangle ABC, if $\frac{1}{a+c}+\frac{1}{b+c}=$$\frac{3}{a+b+c}$ then C is equal to

The given relation can be written as

$\begin{array}{l}\frac{a+b+2c}{(a+c)(b+c)}=\frac{3}{a+b+c}\\ \Rightarrow (a+b+2c)(a+b+c)=3(a+c)(b+c)\\ \Rightarrow (a+b{)}^2+3c(a+b)+2c^2=3\left(ab+ac+bc+c^2\right)\\ \Rightarrow a^2+b^2-c^2=ab\end{array}$

$\begin{array}{ll}\therefore & \cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{ab}{2ab}=\frac{1}{2}\\ \Rightarrow & C={60}^{\circ }\end{array}$