You don't need to put it in "row" or "column" form at all. Any independent subset of a set of vectors will form a basis for the span of the set.
You say that you know that the dimension of the span is two, not three, so you must know that the set is not independent. How did you determine that?
The definition of "independent" is that the only way you can have a linear combination, a_1v_1+ a_2v_2+ a_3v_3= 0, equal to 0, is if all coefficients are equal to 0: a_1= a_2= a_3= 0. Here, that equation becomes
a_1(0, 1, -2)+ a_2(3, 0, 1)+ a_3(3, 2, -3)= (3a_2+ 3a_3, a_1+ 2a_3, -2a_1+ a_2- 3a_3)= (0, 0, 0)
So we must have 3a_2+ 3a_3= 0, a_1+ 2a_3= 0, and -2a_1+ a_2- 3a_3= 0. From the first equation, a_2= -a_3, and, from the second, a_1= -2a_3. Putting those into the third equation, -2(-2a_3)+ (-a_3)- 3a_3= 3a_3- 3a_3= 0 for all a_3 so there exist non-zero solutions. Yes, these vectors are dependent.
Now, since a_1= -2a_3 and a_2= -a_3, our original equation, a_1(0, 1, -2)+ a_2(3, 0, 1)+ a_3(3, 2, -3)= -2a_3(0, 1, -2)- a_3(3, 0, 1)+ a_3(3, 2, -3)= 0 so we can write a_3(3, 0, 1)= -2a_3(0, 1, -2)+ a_3(3, 2, -3) or (3, 0, 1)= -2(0, 1, -2)+ (3, 2, -3). Since (3, 0, 1) can be written as a linear combination of (0, 1, -2) and (3, 2, -3), any linear combination all three (any vector in the span) can be written as a linear combination of those two. Since they independent (one is not a multiple of the other) (0, 1, -2) and (3, 2, 3) are a basis for that span.
(In fact, since we could solve -2a_3(0, 1, -2)- a_3(3, 0, 1)+ a_3(3, 2, -3)= (0, 0, 0) for any one vector as a linear combination of the other two, any pair of the original vectors is a basis.)
Writing these vectors as rows or columns and then row or column reducing may be quicker, depending on the vectors, but you should always keep in mind the basic definitions.
