# If $\Delta ABC$ if $\large\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$ then $C=$

$\begin {array} {1 1} (1)\;90^{\circ} & \quad (2)\;60^{\circ} \\ (3)\;45^{\circ} & \quad (4)\;30^{\circ} \end {array}$

(2) $60^{\circ}$