$\begin{array}{r}x+y+6 z=1 \\ \frac{1}{3} x-\frac{1}{3} y+\frac{2}{3} z=3 \\ \frac{1}{2} x+z=0 \\ (x, y, z)=(\square)\end{array}$

Step 1 ：First, convert the system of equations into matrix form. The coefficients of x, y, and z in each equation will form the matrix 'A' and the constants on the right side of the equations will form the vector 'b'.

Step 2 ：Matrix 'A' and vector 'b' are as follows: \(A = \left[ \begin{array}{ccc} 1 & 1 & 6 \ \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \ \frac{1}{2} & 0 & 1 \end{array} \right]\) and \(b = \left[ \begin{array}{c} 1 \ 3 \ 0 \end{array} \right]\)

Step 3 ：Next, solve for the vector 'x' which contains the values of x, y, and z.

Step 4 ：The solution to the system of equations is \(x = \left[ \begin{array}{c} -5 \ -9 \ 2.5 \end{array} \right]\)

Step 5 ：Finally, the solution to the system of equations is \((x, y, z) = \boxed{(-5, -9, 2.5)}\)