I think you can do it this way:
The first thing to notice is that dim(V)=3 and dim(W)=2. Hence:
Repeating the process for all the standard basis of a 3D space gives the vectors which span W to be and .
However these vectors just span. If we have three vectors which span, and the vector space only has a dimension of 2, one of the vectors must be a linear combination of the others.
The vector is a linear combination of 15 vectors and -10 vectors (these were found using simultaneous equations).
Therefore, upon sifting, we get: