A One-Two-Three Tetrahedron

A tetrahedron $P\text{-}ABC$ is inscribed in a sphere $O$, with $PA : PB : PC = 1 : 2 : 3$. When the lateral surface area of the tetrahedron (the total area of the three faces meeting at $P$) is largest, the volume of the sphere $O$ is $\dfrac{56\sqrt{14}}{3}\pi$. Under these conditions, find the area of triangle $ABC$.