Answer
No solution.
Work Step by Step
Step 1. Set up the augmented matrix of the system of equations:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-1&-2&3&0\\ 0&1&-1&1&1\\ 3&-2&-7&10&2 \end{bmatrix}
\begin{array} \\ \\ \\R_3-3R_1\to R_3\\ \end{array}$
Step 2. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-1&-2&3&0\\ 0&1&-1&1&1\\ 0&1&-1&1&2 \end{bmatrix}
\begin{array} \\ \\ \\R_3-R_2\to R_3 \\ \end{array}$
Step 3. Do the row operations indicated on the right side of the matrix:
$\begin{array} \\A=\\ \end{array}
\begin{bmatrix} 1&-1&-2&3&0\\ 0&1&-1&1&1\\ 0&0&0&0&1 \end{bmatrix}
\begin{array} \\ \\ \\ \\ \end{array}$
Step 4. The last row gives $0-1$ which mean that there is no solution to the system.
