Answer:
Option C: (2, 1, 0)
Explanation:
We are given a system of three equations. Usually, we eliminate a variable from the equations. Solve for the remaining two equations and determine the value of the third equation later.
Here, we are given the choices. So, we can readily substitute each option in the equations. The solution should satisfy all the three equations.
Option A: (3, 1, 0)
Equation (1): 3x + 3y + 6z = 9
⇒ 3(3) + 3(1) + 6(0) = 9 + 3 ≠ 9
Option A does not satisfy the first equation. So, it is eliminated.
Option B: (0, 1, 0)
Substituting in the first equation: 3x + 3y + 6z = 9
⇒ 3(0) + 3(1) + 6(0) = 3 ≠ 9
Option B does not satisfy as well. Hence, it is eliminated.
Option C: (2, 1, 0)
Substituting in 3x + 3y + 6z = 9
⇒ 3(2) + 3(1) + 6(0) = 9
It satisfies Equation 1. For this to be a solution, it should satisfy all the three equations. So, we check for equation 2 and 3.
Equation 2: x + 3y + 2z = 5
2 + 3(1) + 2(0) = 2 + 3 = 5
Equation 3: 3x + 12y + 12z = 18
3(2) + 12(1) + 12(0) = 6 + 12 = 18
Option C satisfies all the equations. So, it is the solution to the system.
On simple substitution, Option D can be eliminated as well.
Hence, the answer.