In a triangle $A B C$, with usual notations, $(a+b+c)(a+b-c)=3 a b$, then $\angle \mathrm{C}=$

Given equation: $(a+b+c)(a+b-c)=3ab$.

Applying the difference of squares identity, $(a+b+c)(a+b-c)= (a+b)^2 - c^2$.

Expanding and substituting gives $a^2 + b^2 + 2ab - c^2 = 3ab$, which simplifies to $a^2 + b^2 - c^2 = ab$.

Using the Law of Cosines: $a^2 + b^2 - c^2 = 2ab \cos C$, so we equate to get $2ab \cos C = ab$.

Dividing both sides by $ab$ (since $a \neq 0$ and $b \neq 0$) gives $\cos C = \frac{1}{2}$, so $\angle C = \frac{\pi}{3}$.

Final answer: $\boxed{C}$