spanning over vectors, but this span has only 1 vector
Solution 1:
H spans (and is a basis for) a 1 dimensional subspace of R$^4$.
Solution 2:
Take $R^3$ and cut your span to $span[2, 0, 3]$. Then we have the vector $2, 0, 3$, which spans multiples of $2, 0, 3$ so we have $n(2, 0, 3)$.
Now for $R^4$, do the same. As Betty mentioned it is one-dimensional.
If we had two lines, each with different one-dimensional span, we would end up getting a plane.