The correct answer is $\boxed{0}$.
To solve the system of linear equations, we can use the elimination method. First, we add the first and second equations together. This gives us $3x_1 + 3x_2 = 2$. Then, we add the third equation to this equation. This gives us $4x_1 + 4x_2 = 4$. Dividing both sides by 4, we get $x_1 + x_2 = 1$.
Now that we know the value of $x_1$, we can substitute it into any of the original equations to solve for $x_3$. Substituting it into the first equation, we get $x_1 + 2x_2 – 2x_3 = 4$. Simplifying, we get $x_2 – 2x_3 = 4$. Adding $2x_3$ to both sides, we get $x_2 = 6$.
Substituting $x_1 = 1$ and $x_2 = 6$ into the third equation, we get $-1 + 6 – x_3 = 2$. Solving for $x_3$, we get $x_3 = 0$.
Therefore, the value of $x_3$ obtained by solving the following system of linear equations is $\boxed{0}$.
Option A is incorrect because $-12$ is not a solution to the system of linear equations.
Option B is incorrect because $-2$ is not a solution to the system of linear equations.
Option C is incorrect because $0$ is a solution to the system of linear equations.
Option D is incorrect because $12$ is not a solution to the system of linear equations.