The regular triangular pyramid means that it has an equilateral base. Let the sidelength of the base be $2\sqrt{3}x$ then the height of the pyramid is $\sqrt{36-x^2}$ and the volume is $$\frac{1}{3}(\sqrt{36-x^2})(3\sqrt{3}x^2)=99$$ $$(\sqrt{36-x^2})(3\sqrt{3}x^2)=297$$ Let $a=x^2$ $$a\sqrt{36-a}=33\sqrt{3}$$ $$a=33, x=\sqrt{33}$$ Thus the perimeter was $6\sqrt{3}x=(6\sqrt{3})(\sqrt{33})=\boxed{18\sqrt{11}}$