What is the result of a classical collision between THREE point particles at the same precise instant? Classical Mechanics is said to be deterministic, a statement that nearly always is followed by that quote from Laplace, something like

If at one time, one knew the positions and velocities of all the
  particles in the universe, the laws of science should enable us to
  calculate their positions and velocities at any other time, past or
  future.

I always scratch my head after hearing/reading that. If 3 or more rigid point particles happen to collide elastically at the same precise instant, is it not impossible to predict the resulting trajectories? If it is possible, how?
 A: Taking the case of point particles and "contact" collisions seriously actually causes trouble even in the two dimensional case: the instantaneous forces are necessarily infinite even if the impulses remain finite.
The solution to that problem--to recognise that all real particles interact via fields over non-zero distances--solves the three particle problem as well. You just integrate the equations of motion (possibly numerically as this may not be easy in closed form).
This isn't necessarly in the 2 particle elastics case because conservation of energy and momentum fully constrain the outcome allowing us to elide this question in a introductory presentation.
A: I have not found any reference on this. The following is my approach for point particles, colliding in the same point of space at the same time (elastic collision). 
Let us consider 2D motion and the particles traveling towards the collision point. If the masses and momenta were equal, it would be natural to assume by symmetry, scatter back with equal momenta, in the same incoming directions, but in opposite sense. 
While if only one particle had different momentum to the other two the symmetry would be along the uneven one. The components of the two equal particles, perpendicular to the symmetry line, should be unchanged (only the sense to opposite). And the components along the symmetry line would be calculated by the balance of the momentum of the incoming particle and the sum of the components of the remaining two (2 times the value of one due to symmetry), which would replicate the case of collision with a particle twice as massive as one of the two equal, whose velocity is reduced by the cosine (because is the component of the original momentum).
As for the case where all momenta and masses are different, I would generalise by balancing each particle with the components of the other two along its direction. This would give three equations, but only two of them would be linearly independent. This would be something similar to the figure (although in such figure the initial momenta are zero and the particles attract each other)

Finally note that the mentioned restriction (2D motion), applies also for 3D collisions, because the center of mass motion occurs in one plane.
Indeed, if we choose the lines of motion of two of the particles, these two lines define a plane. The third line will in general be out of this plane. Let us further define the plane containing the three particles in any moment, as the center of mass plane. The projections of the momenta of each of the particles in this plane does not change in time, since the total momenta and their angles with this plane do not. Even more, the components of momenta perpendicular to this plane are equal for all particles, as this plane moves towards the collision point. Hence the name center of mass plane. (if you prefer, I can provide a mathematical proof)
