Find the solution to the system of equations: \[ \left\{\begin{array}{l} x-2 y+z=-1 \\ y+2 z=-3 \\ x+y+3 z=-6 \end{array}\right. \] \[ \begin{array}{l} x= \\ y= \\ z= \end{array} \]

Step 1 ：We are given a system of linear equations with three variables as follows:

Step 2 ：\[\left\{\begin{array}{l} x-2 y+z=-1 \\ y+2 z=-3 \\ x+y+3 z=-6 \end{array}\right.\]

Step 3 ：We can solve this system using various methods such as substitution, elimination or matrix method. Here, we will use the matrix method.

Step 4 ：First, we represent the system of equations in matrix form. The matrix A represents the coefficients of the variables, and the matrix B represents the constants on the right side of the equations.

Step 5 ：A = \[\begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & 2 \\ 1 & 1 & 3 \end{bmatrix}\] and B = \[\begin{bmatrix} -1 \\ -3 \\ -6 \end{bmatrix}\]

Step 6 ：We then solve for the variables x, y, and z by finding the inverse of matrix A and multiplying it with matrix B.

Step 7 ：The solution is \[\begin{bmatrix} -2 \\ -1 \\ -1 \end{bmatrix}\]

Step 8 ：Thus, the solution to the system of equations is

Step 9 ：\[\begin{array}{l} x= \boxed{-2} \\ y= \boxed{-1} \\ z= \boxed{-1} \end{array}\]