The correct answer is: A. unique solution
To solve a system of equations, we can use the elimination method. In this method, we eliminate one variable at a time by adding or subtracting the equations in a way that cancels out the variable.
For the system of equations $x + 2y + z = 6$, $2x + y + 2z = 6$, and $x + y + z = 5$, we can eliminate $z$ by adding the first two equations together. This gives us $3x + 3y = 12$. We can then eliminate $y$ by subtracting the third equation from the second equation. This gives us $x = 1$. Substituting $x = 1$ into the first equation, we get $1 + 2y + z = 6$. Solving for $y$, we get $y = 2$. Substituting $x = 1$ and $y = 2$ into the third equation, we get $1 + 2 + z = 5$. Solving for $z$, we get $z = 2$.
Therefore, the system of equations has a unique solution $(x, y, z) = (1, 2, 2)$.
Here is a brief explanation of each option:
- Option A: unique solution. This means that there is only one set of values for $x$, $y$, and $z$ that satisfies the system of equations.
- Option B: infinite number of solutions. This means that there are infinitely many sets of values for $x$, $y$, and $z$ that satisfy the system of equations.
- Option C: no solution. This means that there is no set of values for $x$, $y$, and $z$ that satisfies the system of equations.
- Option D: exactly two solutions. This means that there are exactly two sets of values for $x$, $y$, and $z$ that satisfy the system of equations.