That is the proper way of doing it. Pick any two linearly independent solutions of the system Ax=0. Then let B be the matrix whose first two columns are the solutions, and fill the remaining columns with zeros.
You can do the second bit easily if you know the following inequality. If A is an mxn matrix, and B is an nxp matrix, then
rank(A) + rank(B) - n <= rank(AB)
(Try to justify it by drawing a diagram or something, if you don't feel like proving it carefully.)
So in this case,
3 + rank(B) - 5 <= 0
=> rank(B) <= 2