Let be the line of intersection of the planes given by the equationsLet be the line passing through the point and parallel to . Let denote the plane given by the equationSuppose that the line meets the plane at the point . Let be the foot of the perpendicular drawn from to the plane .If the length of the line segment is and the area of is , find the value of 4$$\ell$$^2 A^2.\
[THOUGHT]
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Find the direction vector of line L1 by subtracting the coefficients of the two planes' equations. Direction vector of L1 = (2-1, 3-2, 0-1) = (1, 1, -1)
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Line L2 is parallel to L1 and passes through P(2, -1, 3), so its direction vector is the same as L1's. Parametric equation of L2: (x, y, z) = (2, -1, 3) + t(1, 1, -1)
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Find a point Q on L2 that also lies in plane Π by substituting L2's parametric equations into Π's equation. 2*(2+t) + (-1+t) - 2*(3-t) = 6 Solving for t gives: t = 1/2
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Substitute t=1/2 back into L2's parametric equation to get point Q: (x, y, z) = (2, -1, 3) + 1/2(1, 1, -1) Point Q = (7/2, 1/2, 1/2)
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Find vector PQ: (7/2-2, 1/2+1, 1/2-3) = (3/2, 3/2, -5/2)
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Vector PR is P to R, which is parallel to Π's normal vector (2, 1, -2). Normalizing Π's normal vector gives: n = (2/sqrt(10), 1/sqrt(10), -2/sqrt(10))
So, unit vector in the direction of PR = (2/sqrt(10), 1/sqrt(10), -2/sqrt(10)) Vector PR = (2/sqrt(10)) * |PQ| = (3/2)*sqrt(10)/2
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Area A of triangle PQR is half the cross product of PQ and PR: Cross product: (3/2, 3/2, -5/2) x ((3/2)sqrt(10)/2, (1/2)sqrt(10)/2, -(3/2)sqrt(10)/2) = (15sqrt(10)/4 + 15/4(-2), 9sqrt(10)/4 - 9/4*(-5), 9sqrt(10)/4 + 3/4(-15)) = (21.75, 45.75, -6)
Area A = |cross product / 2| = sqrt((21.75)^2 + (45.75)^2 + (-6)^2) / 2 ≈ 16.39
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Length of PQ: |PQ| = sqrt((3/2)^2 + (3/2)^2 + (-5/2)^2) = sqrt(28.5) ≈ 5.35
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Compute the answer: 4 * (length PQ)^2 * (Area)^2 = 4 * (5.35)^2 * (16.39)^2 ≈ 1,293,544
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