$\left\{\begin{array}{c}2 x+y+3 z=13 \\ x+2 y=11 \\ 3 x+z=10\end{array}\right.$

Step 1 ：We are given the system of linear equations: \[\begin{cases} 2x + y + 3z = 13 \\ x + 2y = 11 \\ 3x + z = 10 \end{cases}\]

Step 2 ：We can represent this system in matrix form as: \[\begin{bmatrix} 2 & 1 & 3 \\ 1 & 2 & 0 \\ 3 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 13 \\ 11 \\ 10 \end{bmatrix}\]

Step 3 ：By solving this matrix equation, we find the solution to be: \[\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ 1 \end{bmatrix}\]

Step 4 ：So, the solution to the system of equations is \(x = 3\), \(y = 4\), and \(z = 1\).

Step 5 ：Final Answer: \(\boxed{x = 3, y = 4, z = 1}\)