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Let P1: x + y + 2z - 3 = 0 and P2: x - 2y + z - 4 = 0 be two planes. Let A(1, 3, 4) and B(3, 2, 7) be two points. Plane P3 passes through the line of intersection of P1 and P2 such that the length of projection of segment AB on P3 is maximised. If the equation of P3 is ax + by + cz + 1 = 0 (where a, b, c are integers), which of the following is correct?

1. a + b + c = 4
2. a + 2b + c = 7
3. a - b + c = 0
4. a + 2b + c = 5

Correct answer: a + b + c = 4

Solution

Projection of AB on P3 is maximised when AB's component along normal of P3 is zero, i.e., AB is perpendicular to the normal of P3. AB = B - A = (2, -1, 3). Family: P3: (1+lambda)x + (1-2*lambda)y + (2+lambda)z - (3+4*lambda) = 0. Normal: n = (1+lambda, 1-2*lambda, 2+lambda). Condition n.AB = 0: 2*(1+lambda) + (-1)*(1-2*lambda) + 3*(2+lambda) = 0. 2+2*lambda - 1+2*lambda + 6+3*lambda = 0. 7 + 7*lambda = 0. lambda = -1. P3: (1-1)x + (1+2)y + (2-1)z - (3-4) = 0. 0*x + 3y + z + 1 = 0. So 3y + z + 1 = 0. Compare with ax+by+cz+1=0: a=0, b=3, c=1. a+b+c = 0+3+1 = 4. Answer: a+b+c = 4.

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