Answer
No solution.
Work Step by Step
Step 1. Establish the augmented matrix of the system:
$\begin{vmatrix} 2 & 1 & 3 & 9 \\-1 & 0 & -7 & 10\\3 & 2 & -1 & 4 \end{vmatrix} \begin{array}(-R_2\leftrightarrow R_1 \\.\\.\\ \end{array}$
Step 2. Multiple row 2 by $-1$ and exchange with row 1:
$\begin{vmatrix} 1 & 0 & 7 & -10\\2 & 1 & 3 & 9 \\3 & 2 & -1 & 4 \end{vmatrix} \begin{array} ..\\ R_2- 2R_1\to R_2, \\R_3-3R_1\to R_3\end{array}$
Step 3. Use subtract twice row1 from row2 and subtract $3R_1$ from row 3:
$\begin{vmatrix} 1 & 0 & 7 & -10\\0 & 1 & -11 & 29 \\0 & 2 & -22 & 34 \end{vmatrix} \begin{array}..\\.\\R_3-2R_2\to R_3\end{array}$
Step 4. Subtract twice of row 2 from row 3:
$\begin{vmatrix} 1 & 0 & 7 & -10\\0 & 1 & -11 & 29 \\0 & 0 & 0 & -24 \end{vmatrix}$
Step 5. The third row of the matrix shows $0=-24$ indicating that the system has no solution.
