first of all, you didn't "finish" row-reduction, because you only reduced your matrix to upper-triangular form (and not row echelon, or reduced row echelon form, which makes the solutions more transparent).
but look, your last row means that:
0x + 0y - 3z = -3 that is:
-3z = -3
z = 1 <--not a parameter, a VALUE.
your second row means:
0x + 2y + 6z = 2 that is:
2y + 6z = 2
but we already know that z = 1! so this becomes:
2y + 6 = 2
2y = -4
y = -2 <---again, not a parameter, an actual number.
finally, your top row means:
x + 4y + 11z = 7
since y = -2, ans z = 1, this is:
x - 8 + 11 = 7
x + 3 = 7
x = 4
the "other" question you ask, how do you know "when to use parameters?" depends on the RANK of your original matrix.
row-reduction is one way we find out the rank, after row-reduction, it is the number of non-zero rows remaining. in this case, the rank of the original matrix is 3.
this means several things, all of which are equivalent:
1) the matrix A in your original problem is invertible
2) the matrix A is non-singular (the null space is ONLY the 0-vector (0,0,0)) that is A(x,y,z) = (0,0,0) implies x = y = z = 0.
3) the column space of A has dimension 3 (it is thus ALL of R3).
4) the columns of A are linearly independent
5) the equation A(x,y,z) = (7,9,4) has a UNIQUE solution
6) the determinant of A is non-zero
7) we can use Cramer's Rule to solve the system of equations