The equation of the plane through the intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x−y+z=0 is:

The equation of the plane through the intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x−y+z=0 is:

• A. x − z + 2 = 0
• B. x − y + z = 1
• C. y − z = 0
• D. x − z = 2

Answer: A) x − z + 2 = 0

Explanation: Plane equation: (x+y+z−1) + λ(2x+3y+4z−5) = 0. Normal vector: (1+2λ)i + (1+3λ)j + (1+4λ)k. It's perpendicular to x−y+z=0, so 1(1+2λ) − 1(1+3λ) + 1(1+4λ) = 0 → 1 + 3λ = 0 → λ = −1/3. Substituting λ: 3(x+y+z−1) − 1(2x+3y+4z−5) = 0 → x − z + 2 = 0.