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Two lines x - 1 2 = y - 2 3 = z - 3 4 and x - 4 5 = y - 1 2 = z 1 intersect at a point P . If the distance of P from the plane 2 x - 3 y + 6 z = 7 is λ units, then the value of 49 λ is equal to

Correct answer

84

Step-by-step solution

Point P must lies on both the given lines From 1 s t line coordinate of point P 2 λ + 1,3 λ + 2,4 λ + 3 This must satisfy 2 n d line i.e. 2 λ - 3 5 = 3 λ + 1 2 = 4 λ + 3 1 ⇒ 4 λ - 6 = 15 λ + 5 ⇒ λ = - 1 So, coordinates of point P are - 1 , - 1 , - 1 Hence, the distance of P from the plane is λ = - 2 + 3 - 6 - 7 7 = 12 7 units

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