Following up on JSE's answer, this arrangement is, up to linear change of coordinates, the projectivization of the braid arrangement of rank three, so the fundamental group of the complement is the quotient of the pure braid group (of the plane) on four strands by the infinite cyclic subgroup generated by the full twist (= the product of the usual generators), which is central. You can see this as follows: the four points are labelled x,y,z,w; the requirement that they are distinct means you throw away the hyperplanes x=y, x=z, x=w, etc. The complement of these six hyperplanes in C^4 is homotopy equivalent to the x=0 slice: the homotopy is given by sliding points parallel to the line x=y=z=w, which lies in all the hyperplanes. This gives the arrangement of six planes in C^3 with equations y=0,z=0,w=0,y=z,y=w,z=w. projectively, once you throw away y=0 you can set y=1 in the other five equations, and you get your arrangement.