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Let P1: x + y + 2z - 3 = 0 and P2: x - 2y + z = 4 be two planes, and let A(1,3,4) and B(3,2,7) be two points in R³. The plane P3 passes through the line of intersection of P1 and P2. Among all such planes P3, find the one on which the projection of segment AB is greatest. If the equation of this plane is ax + by + cz + 1 = 0 (where a, b, c are integers), find a + b + c.

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

Vector AB = B - A = (2,-1,3). Any plane through the line of intersection of P1 and P2 has the form P1 + lambda*P2 = 0: (1+lambda)x + (1-2*lambda)y + (2+lambda)z + (-3-4*lambda) = 0. Normal to P3: n = ((1+lambda), (1-2*lambda), (2+lambda)). Projection of AB on P3 is maximized when n is perpendicular to AB: n. AB = 0. n. AB = 2(1+lambda) + (-1)(1-2*lambda) + 3(2+lambda) = 2+2*lambda - 1+2*lambda + 6+3*lambda = 7+7*lambda = 0 => lambda = -1. Plane: (1-1)x + (1+2)y + (2-1)z + (-3+4) = 0 => 0*x + 3*y + 1*z + 1 = 0 => 3y + z + 1 = 0. So a=0, b=3, c=1 -> a+b+c = 4. Hmm that gives 4. Let me recheck lambda=-1: (1+(-1))=0, (1-2(-1))=3, (2+(-1))=1, (-3-4(-1))=1. Equation: 0x + 3y + z + 1 = 0. a+b+c = 0+3+1 = 4. Answer = 4.

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