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Consider planes P1: x + 2y + z - 1 = 0 and P2: x - y + 2z + 1 = 0. A plane P contains the line of intersection of P1 and P2. Given that P is perpendicular to the plane x + y + z = 5, and the equation of P is Ax - By + Cz + 3 = 0 where A, B, C are positive integers, find all integers greater than which (A + B + C) lies.

1. 3
2. 5
3. 7
4. 9

Correct answer: 7

Solution

Family of planes through line of intersection of P1 and P2: P = P1 + lambda*P2 = 0 (x + 2y + z - 1) + lambda(x - y + 2z + 1) = 0 (1+lambda)x + (2-lambda)y + (1+2*lambda)z + (-1+lambda) = 0 P perpendicular to x + y + z = 5 means their normals are perpendicular: n_P. nₓ+y+z = 0 (1+lambda)(1) + (2-lambda)(1) + (1+2*lambda)(1) = 0 1 + lambda + 2 - lambda + 1 + 2*lambda = 0 4 + 2*lambda = 0 lambda = -2 Substitute lambda = -2: (1-2)x + (2+2)y + (1-4)z + (-1-2) = 0 -x + 4y - 3z - 3 = 0 Multiply by -1: x - 4y + 3z + 3 = 0 So A=1, B=4, C=3, constant=3. Check: A,B,C > 0. A+B+C = 1+4+3 = 8. 8 is greater than 3, 5, and 7. So A+B+C > 7.

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