That's not right. The row space has dimension 3, not two.

The system of equations for what? If you are looking for the null space, why not use the original matrix?So that gives me the system of equations:

{x1 + 9x3 + 3x4 = 0

{x2 + 5x3 = 0

Any (x,y,z,u) in the null space must satisfy x- 2y- z+ 3u= 0, 4x+7y+ z+ 12u= 0, 3x- 5y+ 2z+ 9u= 0. Adding the first two equations gives 5x+ 5y+ 15u= 0 or x+ y+ 15u= 0, eliminating z. Subtracting the twice the first equation from third equation gives x- y+ 3u= 0, also eliminating z. Adding those two equations gives 2x+ 18u= 0, or x= -9u, eliminating y. Since that is as far as we can go, the null space has dimension 1. We can choose u to be any non-zero number and solve for x, y , and z to get a basis: If u= 1, x= -9 so x-y+ 3u= 0 becomes -9- y+ 3= 0 or y= 6. Then x- 2y- z+ 3u= 0 becomes -9- 12+ z+ 3= 0 or z= 18. {(-9, 6,18, 1)} is a basis for the null space.Now I'm lost. I can't seem to manipulate this to get the null space. Any hints?

Yes, you could do that by row-reducing the matrix which would be the same as using basis vectors for the row space but the point is you did that incorrectly!

To find a basis for the column space, find a basis for the row space of the transpose:

Add twice the first row to the second , add the first row to the third and subtract 3 time the first row from the third to get

Swap the second and third rows, then subtract 3 times the (new) second row from the (new) third row to get

A basis for the column space is {(1, 4, 3), (0, 5, 5), (0, 0, -14)}

The row space has dimension 2, the null space dimension 1, and the column space dimension 3. Notice that the sum of the row space dimension and the nullity is 3, the number of rows. The sum of the column space dimension and nullity is 4, the number of columns.