There are many ways to approach this sort of question, based on how much one depends on mechanical approaches vs. intuitive approaches.
Perhaps the most mechanical way of approaching this question is to quite literally Gram-Schmidt these guys to death (as this removes redundant vectors, i.e. linearly dependent vectors, i.e. vectors that are already spanned in the vectors already considered). For example, after one performs G-S on the first set of vectors, we are only left with 2 basis vectors (the third is obliterated). Continuing this, we obliterate the second set of vectors with these two basis vectors. (We note that the second set of vectors, considered separately, are linearly independent - and so also span a space of 2 dimensions).
But what does this accomplish? It's just a mechanical way of explicitly writing each vector in terms of other vectors. So we literally show that anything written in one basis can be written in the other. That's not so exciting.
Suppose we wanted to do it differently still: we can quickly see that the dimension of the space spanned by the first set is 2, and the dimension of the space spanned by the second set is 2. These are so called linear spaces, which is handy. This means that if we find 2 linearly independent vectors contained in both spaces, they are the same. While in one respect, we could just choose the two first vectors and see if they're in the second space, this also means that we could choose any linear combination of the first two vectors as well. So it's slightly more general.
Differently still: each linear subspace is 2 dimensional, i.e. a plane through the origin (it's a linear space, so it contains the origin). You could find the equation of the plane generated by the first two vectors and compare it to the plane generated by the two vectors in the second set. They are the same (multiples of each other). How does one do this? Taking cross products! (if you know them - it's very very simple).
Differently still: after realizing that each subspace is 2 dimensional, throw all 5 vectors in a matrix and row reduce. If 2 are left, they're the same. If there are 3 or more, then they are not the same. Along the same lines, one could (though should not, to be honest) proceed with determinants. Form a matrix whose first and second rows are the vectors from the first set, and whose third row is the first vector of the second set. Form another whose third row is the second vector of the second set. Taking determinants, we will see that both determinants are zero! This means that the 3-dimensional volume of these two parallelopipeds is zero, i.e. that they lie flat! If they lie flat, their sides must be linearly dependent, and since both vectors of the second set are dependent in the first set, they span the same subspace.
Differently still: find a vector not spanned in the first set, find the component orthogonal to the first subspace, and dot this orthogonal component with each vector in the second set. You will get 0 both times, meaning that the two subspaces have the same orthogonal complement, and therefore they are the same. Alternatively, take the cross product of the two vectors in the first set and dot the result with each vector in the second set. You will get 0 again, and this does not bear the burden of finding orthogonal complements in any witty or projective fashion.
Differently still: do it heuristically! By rolling a fair 24-sided die (called a deltoidal icositetrahedron) repeatedly, generate a random set of about 9 different points in 3 space. Find the line of best fit and project onto these two spaces. You will get the same projections! After doing this a couple of times, you can expect this to always work! Afterwords, show that as these 9 points are always on lattice points, at least 1 of the 36 segments joining these 9 points contain a lattice point as well. It will sharpen your skills with pigeonholing ideas.
Differently still, and finally: guess. When I TA'd linear algebra, my students would throw everything they could into matrices and row-reduce them, write a few things about rank and nullity, and solve a linear system of equations (whether asked for or not) on every question. Literally, even this one. When they realized that's not what I asked for, they would write a few illegible lines of work, draw a big $\Longrightarrow$, and say something like "Clearly, they do not span the same space" or "Obviously, they span the same space." For some other TAs, they'd give partial credit. It at least gave me a laugh.
Seriously though, if you would like to explain any of these further, let me know. They all really do work. And yes, it was quite a yarn. But I've been gone for a couple of weeks, and I had to say hello somehow!