Solution: We want to solve the following system using row-reduction: \[ \begin{array}{ccc} \phantom{-}\frac{1}{3}x_1 + 2x_2 - 6x_3 & = & 0 \\[1ex] -4x_1 \phantom{+ 2x_2} + 5x_3 & = & 0 \\[1ex] -3x_1 + 6x_2 - 13x_3 & = & 0 \\[1ex] -\frac{7}{3}x_1 + 2x_2 - \frac{8}{3}x_3 & = & 0. \end{array} \] Consider the augmented matrix. \begin{align*} \begin{pmatrix} \frac{1}{3} & 2 & -6 & 0 \\ -4 & 0 & 5 & 0 \\ -3 & 6 & -13 & 0 \\ -\frac{7}{3} & 2 & -\frac{8}{3} & 0 \end{pmatrix} & \xrightarrow{ \begin{matrix} R_1 \leftrightarrow 3 R_1 \\ R_3 \leftrightarrow 3 R_3 \\ \end{matrix} } \begin{pmatrix} 1 & 6 & -18 & 0 \\ -4 & 0 & 5 & 0 \\ -3 & 6 & -13 & 0 \\ -7 & 6 & -8 & 0 \end{pmatrix} \\[1ex] & \xrightarrow{ \begin{matrix} R_2 \rightarrow R_2 + 4R_1 \\ R_3 \rightarrow R_3 + 3R_1 \\ R_4 \rightarrow R_4 + 7R_1 \end{matrix}} \begin{pmatrix} 1 & 6 & -18 & 0 \\ 0 & 24 & -67 & 0 \\ 0 & 24 & -67 & 0 \\ 0 & 48 & -134 & 0 \end{pmatrix} \\[1ex] & \xrightarrow{ \begin{matrix} R_2 \leftrightarrow \frac{R_2}{24} \end{matrix} } \begin{pmatrix} 1 & 6 & -18 & 0 \\ 0 & 1 & -\frac{67}{24} & 0 \\ 0 & 24 & -67 & 0 \\ 0 & 48 & -134 & 0 \end{pmatrix} \\[1ex] & \xrightarrow{ \begin{matrix} R_3 \rightarrow R_4 - 24 R_2 \\ R_1 \rightarrow R_1 - 6R_2 \\ R_4 \rightarrow R_4 - 48 R_2 \end{matrix} } \begin{pmatrix} 1 & 0 & -\frac{5}{4} & 0 \\ 0 & 1 & -\frac{67}{24} & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \end{align*}

Thus, the RREF of the augmented matrix is: \[ \begin{pmatrix} 1 & 0 & -\frac{5}{4} & 0 \\ 0 & 1 & -\frac{67}{24} & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \] Since the third and fourth rows are all zeros, we have two pivot equations: \begin{align*} x_1 - \frac{5}{4}x_3 &= 0, \\ x_2 - \frac{67}{24}x_3 &= 0. \end{align*} Solving for leading variables: \begin{align*} x_1 = \frac{5}{4}x_3 \quad \text{ and } \quad x_2 = \frac{67}{24} x_3, \\ \end{align*} Letting $x_3 = t$, the solution set is \begin{align*} \mathcal{S} = \left\{t \begin{pmatrix} 5 \\ 67 \\ 24 \end{pmatrix}: t \in \mathbb{R} \right\} \end{align*}