T(x,y) = ((2x-(y/2)), -x+3y, 2y)
I have found the kernel which is 0 how do we find the nullity of this?
Therefore it is NOT in the kernel of the linear transformation.
But, since you said "I have found the kernel which is 0" in your first post, I think you were asking in general, not for this particular transformation. It is not possible for the kernel of a linear transformation to be a single non-zero vector!
The kernel of a linear transformation, A, is the set of all vectors, v, such that Av= 0. That is not true for (2, 0). Further, if the kernel is not "trivial", that is, if Av= 0 for any non-zero v, it cannot consist of a single vector! If v is a non-zero vector such that Av= 0 then A(xv)= xAv= 0 for any number x so there must exist an infinite number of vectors in the kernel. The kernel of a linear transformation is a subspace, possibly trivial, which is why it has a dimension.