(a,b,c,d) is in the subspace if a,b,c, and d satisfy the system:
a - 3b + d = 0
c - 2d = 0
So if d = parameter t, then c = 2t
And if b = parameter r, then a = 3b - d = 3r - t
So the solution set is
(a,b,c,d) = (3r - t, r, 2t, t)
= (3r, r, 0, 0) + (-t, 0, 2t, t)
= r(3,1,0,0) + t(-1,0,2,1)
and so the basis is { (3,1,0,0) , (-1,0,2,1) } which is a subspace of dimenion 2 since there are two basis vectors.