Solve the system. \[ \begin{array}{r} x-y+2 z=8 \\ 2 x+z=5 \\ x+3 y+z=11 \end{array} \] A. $\{(0,2,5)\}$ B. $\{(5,2,0)\}$ C. $\{(5,0,2)\}$ D. $\varnothing$

Step 1 ：This is a system of linear equations with three variables. We can solve this system using various methods such as substitution, elimination or matrix method. Here, we will use the matrix method.

Step 2 ：First, we write the system of equations in matrix form. The matrix A is formed by the coefficients of the variables in the system of equations, and the matrix b is formed by the constants on the right side of the equations. So we have: \[A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 1 & 3 & 1 \end{bmatrix}\] and \[b = \begin{bmatrix} 8 \\ 5 \\ 11 \end{bmatrix}\]

Step 3 ：We then solve the system by finding the inverse of matrix A and multiplying it with matrix b. The solution is the vector \[x = \begin{bmatrix} 0 \\ 2 \\ 5 \end{bmatrix}\]

Step 4 ：Final Answer: The solution to the system of equations is \(\boxed{\{(0,2,5)\}}\)