First, some background and notation. In the FE method, you set up the following matrix system, Ku=f. You are looking for "u" given the K and f. The reason you need to put in at least a 3-2-1 support scheme (AKA "rigid body constraint") is that without the 3-2-1 scheme, the K matrix is singular---in 3D, there are at least 6 equations that are dependent on there other N-6 equations in the Ku=f equations (after all, the Ku=f is really just a handy notation and solution scheme for solving N equations, is it not?). That's the 25 cent answer why you need to do this 3-2-1 rigid body contraint.
Perhaps each FE software does it differently? Here's what I would do to constrain the K matrix in Ku=f--in particular, how is this 3-2-1 constraint typically handled? You are constraining 6 degrees of freedom and forcing them to be zero. Your solution is 'u', the displacement vector. You are saying 6 displacements are forced to be zero (remember the displacements "u" is 3 displacements, on each in direction x, y and z, for each node--if you have "M" nodes, you have 3*M=N degrees of freedom). You have N total degrees of freedom (DOFs) in the K matrix; say as part of your 3-2-1 scheme, you are saying displacement no. 383 is zero. You cause your K matrix to be constrained for no. 383 equal to zero by putting a bunch of zeroes on row 383, with a zero in the "f" vector for the f(383). Now put a "1" (one) in the k(383,383)--that is, the diagonal element of matrix K in the 383rd row, 383rd column. If you wrote out the 383rd equation, it would look like u(383)=0. And that's what you should 'compute' for u(383) when you do the inversion, u=K(-1)*f. So there is no 'reaction force' other than zero for this 383rd element of the "f" vector. In fact, if you compute a reaction force much bigger than zero for f(383), then your body is probably not properly constrained. (remember that the "f" vector is a force vector, with three components, one each in the x,y and z directions, at each node).
In my explanation, the "3-2-1" scheme is not the same as a symmetry constraint. With a symmetry constraint, of course you DO expect forces in the direction perpendicular to the edge you are constraining with symmetry constraint.