Answer
No solution.
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-2&3&-2\\ 2&-1&1&2\\ 2&-7&11&-9 \end{bmatrix}
\begin{array} \\ \\R_2-2R_1\to R_2\\-R_3+2R_1\to R_3\\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-2&3&-2\\ 0&3&-5&6\\0&3&-5&5 \end{bmatrix}
\begin{array} \\ \\ \\ \\ \end{array}$
Step 3. Take the difference of the last two rows to get $0=1$ which indicates that there is no solution to the system.
