Let P be the foot of the perpendicular drawn from the point Q(4, -4, 3) to the line L₁: x-1 2 = y+1 1 = z 2 . A plane passes through P and contains the line L₂: x-2 2 = y-1 1 = z -1 . If d is the perpendicular distance from the point R(4, -

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Let P be the foot of the perpendicular drawn from the point Q(4, -4, 3) to the line L₁: x-1 2 = y+1 1 = z 2 . A plane passes through P and contains the line L₂: x-2 2 = y-1 1 = z -1 . If d is the perpendicular distance from the point R(4, -2, 1) to the plane , then the value of 5d^2 is equal to

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Correct answer: 28

Any point on the line L₁ can be taken as A(2t+1, t-1, 2t) . The vector QA = (2t-3, t+3, 2t-3) . Since QA is perpendicular to L₁ , its dot product with the direction vector (2, 1, 2) is zero: 2(2t-3) + 1(t+3) + 2(2t-3) = 0 4t - 6 + t + 3 + 4t - 6 = 0 9t - 9 = 0 t = 1 Thus, the foot of the perpendicular is P(3, 0, 2) . The plane contains the point P(3, 0, 2) and the line L₂ . A point on L₂ is B(2, 1, 0) . The vector BP = (3-2, 0-1, 2-0) = (1, -1, 2) . The direction vector of L₂ is d₂ = (2, 1, -1) . The normal vector

Let P be the foot of the perpendicular drawn from the point Q(4, -4, 3) to the line L₁: x-1 2 = y+1 1 = z 2 . A plane passes through P and contains the line L₂: x-2 2 = y-1 1 = z -1 . If d is the perpendicular distance from the point R(4, -2, 1) to the plane , then the value of 5d^2 is equal to

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