You get exactly the same result. That matrix row reduces to
so that (x, y, z, u) will be in the null space if and only if x+ y- z+ 2u= 0, z+ u= 3, 6u= 0. Those last two equations give z= u= 0 and the first equation reduces to x+ y= 0 or x= -y. Every vector in the null space is (x, -x, 0, 0)= x(1, -1, 0, 0). The null space is one-dimensional and {(1, -1, 0, 0)} is a basis ((-1, 1, 0, 0) is a multiple of that so it would also be a basis vector.