# If all the variables of a matrix each have their own leading ones and the last row is all zeros is it still a unique solution?

The original augmented matrix is $$\begin{pmatrix} \begin{array}{ccc|c} 3 & 3 & 6 & 6 \\ 2 & 2 & 2 & 0 \\ -3 & -3 & -5 & -4 \\ -2 & -1 & -1 & 2 \\ \end{array} \end{pmatrix}$$ and the reduced row echelon form is $$\begin{pmatrix} \begin{array}{ccc|c} 1 & 0 & 0 & -6 \\ 0 & 1 & 0 & 12 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \\ \end{array} \end{pmatrix}$$

Is the unique solution $$$$\left (\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ \end{array}\right ) \ = \left( \begin{array}{c} -6 \\ 12 \\ -2 \\ \end{array} \right)$$$$ or is it an infinite number of solutions?

Yes. If you have more equations than you do unknowns, than any subsequent rows in the matrix should be a zero row, since you can always reduce them using the leading ones from the above rows. If there were more than one zero rows in your RREF, then you would have infinitely many solutions, since then $$x_3$$ would be free.
Here the common rank is the maximal possible rank – $$3$$, so the subspace of solutions has codimension $$3$$. In a $$3$$-dimensional space, this means it has dimension $$0$$, i.e. it is a single point.