Answer
$\{(5,3,6)\}$
Work Step by Step
Use Gauss-Jordan to perform the row operations given, we have:
$\begin{bmatrix} 3 & -4 & 2 & 15 \\ 2 & -1 & 1 & 13 \\ 1 & 2 & -1 & 5 \end{bmatrix} \begin{array} .R1\leftrightarrow R3\\.\\.\\ \end{array}$
$\begin{bmatrix} 1 & 2 & -1 & 5\\ 2 & -1 & 1 & 13 \\ 3 & -4 & 2 & 15 \end{bmatrix} \begin{array} .\\2R1-R2\to R2\\3R1-R3\to R3\\ \end{array}$
$\begin{bmatrix} 1 & 2 & -1 & 5\\ 0 & 5 & -3 & -3 \\ 0 & 10 & -5 & 0 \end{bmatrix} \begin{array} .\\.\\R3-2R2\to R3\\ \end{array}$
$\begin{bmatrix} 1 & 2 & -1 & 5\\ 0 & 5 & -3 & -3 \\ 0 & 0 & 1 & 6 \end{bmatrix} \begin{array} .R1+R3\to R1\\R2+3R3\to R2\\.\\ \end{array}$
$\begin{bmatrix} 1 & 2 & 0 & 11\\ 0 & 5 & 0 & 15 \\ 0 & 0 & 1 & 6 \end{bmatrix} \begin{array} .\\R2/5\to R2\\.\\ \end{array}$
$\begin{bmatrix} 1 & 2 & 0 & 11\\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 6 \end{bmatrix} \begin{array} .R1-2R2\to R1\\.\\.\\ \end{array}$
$\begin{bmatrix} 1 & 0 & 0 & 5\\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 6 \end{bmatrix} \begin{array} .\\.\\.\\ \end{array}$
Thus the solution set is $\{(5,3,6)\}$
