Question:

Step 1: Set up the family of planes containing the two given planes. Any plane containing the line of intersection of planes P_1: 2x - 5y + z = 3 and P_2: x + y + 4z = 5 can be written as: P_1 + λP_2 = 0, or (2x - 5y + z - 3) + λ(x + y + 4z - 5) = 0 Expanding: (2 + λ)x + (-5 + λ)y + (1 + 4λ)z - (3 + 5λ) = 0 Step 2: Apply the parallelism condition. For this plane to be parallel to x + 3y + 6z = 1, their normal vectors must be proportional: Normal of our plane: (2 + λ, -5 + λ, 1 + 4λ) Normal of given plane: (1, 3, 6) For proportionality: (2 + λ)/1 = (-5 + λ)/3 = (1 + 4λ)/6 Step 3: Solve using the first two ratios. (2 + λ)/1 = (-5 + λ)/3 3(2 + λ) = -5 + λ 6 + 3λ = -5 + λ 2λ = -11 λ = -11/2 Step 4: Verify with the third ratio. (2 + λ)/1 = (1 + 4λ)/6 6(2 - 11/2) = 1 + 4(-11/2) 6(-7/2) = 1 - 22 -21 = -21 ✓ Step 5: Substitute λ = -11/2 into the plane equation. (2 - 11/2)x + (-5 - 11/2)y + (1 - 22)z - (3 - 55/2) = 0 (-7/2)x + (-21/2)y - 21z + 49/2 = 0 Multiply by -2/7: x + 3y + 6z - 7 = 0 x + 3y + 6z = 7 ∴ Answer: c