<p>The plane through the intersection of the planes \(x + y + z = 1\) and \(2x + 3y - z + 4 = 0\) and parallel to \(y\)-axis also passes through the point:</p>
Step-by-Step Solution
Key Concept: A plane parallel to the y-axis has normal vector with zero y-component. Use the family of planes through the intersection of two given planes: (x+y+z-1) + λ(2x+3y-z+4) = 0, then impose the parallelism condition to find λ.
Step 1: Use the family of planes through the intersection of the two given planes: (x + y + z - 1) + λ(2x + 3y - z + 4) = 0 Expanding: (1+2λ)x + (1+3λ)y + (1-λ)z + (4λ-1) = 0 Step 2: For the plane to be parallel to the y-axis, the normal vector must be perpendicular to the y-axis direction (0,1,0). This means the coefficient of y must be zero: 1 + 3λ = 0 ⟹ λ = -1/3 Step 3: Substitute λ = -1/3: (1 - 2/3)x + 0·y + (1 + 1/3)z + (-4/3 - 1) = 0 (1/3)x + (4/3)z - 7/3 = 0 Simplifying: x + 4z - 7 = 0 Step 4: Check which point satisfies x + 4z = 7. The plane passes through points where x + 4z = 7 with any value of y. ∴ Answer: D
Correct Answer: D