@Cacks

You need to integrate over the whole surface onto which a pressure is applied, which would lead to solution "A".
A net non-zero sum will emerge when one side applies more torque than the other in average. At least that's for a real hand for which the palm would apply more pressure than the gripping fingers on the other side for instance.

Also I don't know exactly what your inputs are but supposing that you get sets of particles onto which a given "force" is applied I would try to make sure the said "force" is scaled "per particle" = equivalent to a discrete pressure. Otherwise for each set it would be necessary to divide by the related number of particles to avoid artificially multiply the action.

Lots of guessing on my part here, never encountered such an issue TBH!

You should have a look about physics, and mainly the fundamental principle of dynamics, not the one from Newtons with vectors, but the one with torsors (extensions of Newton's laws which deal about moments). You will apply a force where the hand is, and will transport this resulting torsor at the center of rotation. This will give you the torsor of forces at this center, which should have a resultant null and a non null moment.

'r' is the distance from the mass element to the axis of rotation, and the r segment is therefore perpendicular to the axis of rotation (top figure at: https://en.wikipedia.org/wiki/List_of_moments_of_inertia)

In 3D the centre of rotation is actually undefined, and only an axis is to be considered. Also the point where the r segment meets the axis is purely geometrical and does not need to be part of the rotating object if that can help?
 

Yes the axis itself is a line and as such must be fully defined by either a point and a direction or 2 distinct points indeed.
With 1 degree of freedom (= the axis is mechanically constrained) these points can be anywhere provided they geometrically belong to the line.

(with more degrees of freedom, typically 3 for a free solid, then there can be an infinity of different axes. However they all pass through the object's center of mass, which makes it a particular point for rotations too.)