In fact, because you have 5 variables with 3 equations you will need at least 5- 3= 2 such vectors and the equations are dependent, up to 4. And I see now that math2009 has shown that they are NOT independent. 3 vectors are required
yes, you could row-reduce a matrix but I would recommend this: use the three equations to solve for three of the "X"s in terms of the other two, say solve for X3, X4, and X5 in terms of X1 and X2. Then setting X1= 1, X2= 0 will give you one vector in that set and setting X1= 0, X2= 1 will give you another. The choices of "1, 0" then "0, 1" ensure those vectors are independent so they will span the set (in fact they form a basis for the subspace).