Question:

The given system is:<br>2x + y + z = 1<br>x - 2y + z = 2<br>3x - y + 2z = 3<br>Adding the first two equations: 3x - y + 2z = 3, which is the same as the third equation. Thus, the system is dependent and has infinite solutions.<br>Let z = \lambda. Then:<br>2x + y = 1 - \lambda<br>x - 2y = 2 - \lambda<br>Solving for x and y:<br>Multiply the second by 2: 2x - 4y = 4 - 2\lambda<br>Subtracting: 5y = -3 + \lambda => y = (\lambda - 3)/5. This does not match the options directly, let's re-evaluate.<br>Actually, solving the system gives x = (3\lambda-1)/5, y = (\lambda-3)/5, z = \lambda. Wait, let's re-check the options. The options suggest a different parameterization. If we set x = 3\lambda-1, then y = -\lambda, z = -5\lambda+3. Checking this in the equations: 2(3\lambda-1) + (-\lambda) + (-5\lambda+3) = 6\lambda-2-\lambda-5\lambda+3 = 1. Correct. (3\lambda-1) - 2(-\lambda) + (-5\lambda+3) = 3\lambda-1+2\lambda-5\lambda+3 = 2. Correct. 3(3\lambda-1) - (-\lambda) + 2(-5\lambda+3) = 9\lambda-3+\lambda-10\lambda+6 = 3. Correct. Thus (B) is correct. Since \lambda is arbitrary, replacing \lambda with -\lambda gives (-3\lambda-1, \lambda, 5\lambda+3), so (A) is also correct. Checking (C): x-3y+1 = (3\lambda-1)-3(-\lambda)+1 = 6\lambda \neq 0. So (C) is incorrect. Checking (D): 5x+3z = 5(3\lambda-1)+3(-5\lambda+3) = 15\lambda-5-15\lambda+9 = 4 \neq 1. So (D) is correct.