v1 = (2,1,-1), v2 = (-m,-1,3), v3 = (-3,2,m+1), v4=(1,2,1)
Find the values of M for which {v1,v2,v3,v4} spreads R^3
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

We'll start by finding the values of m for which {v1,v2,v4} spreads a plane.
We'll define the line that's spread by v1 and v4:
S = {X*v1 + v4 | X is real}

We need to find a value of X for which X*v1+v4 = v2
and more specifically:

  | 2x + 1 = -m
 <  x  + 2 = -1
  | -x + 1 = 3

The second and third equation of the system show that there is no x that can satisfy the condition, and therefor (v1,v2,v4} spreads a plane for every value of m.