Let $ L $ be a line obtained from the intersection of two planes $ x+2 y+z=6 $ and $ y+2 z=4 $. If point $ P(\alpha, \beta, \gamma) $ is the foot of perpendicular from $ (3,2,1) $ on $ L $, then the value of $ 21(\alpha+\beta+\gamma) $ equals:

Question

Let $ L $ be a line obtained from the intersection of two planes $ x+2 y+z=6 $ and $ y+2 z=4 $. If point $ P(\alpha, \beta, \gamma) $ is the foot of perpendicular from $ (3,2,1) $ on $ L $, then the value of $ 21(\alpha+\beta+\gamma) $ equals:, $ 102 $, $ 142 $, $ 68 $, $ 136 $

MARKS App · Accepted Answer

x + 2 y + z = 6 y + 2 z = 4 × 2 x - 3 z = - 2 ⇒ x = 3 z - 2 ⇒ y = 4 - 2 z x + 2 3 = z y - 4 - 2 = z ⇒ line of intersection of two planes is x + 2 3 = y - 4 - 2 = z = λ (Let) ∵ A P ⊥ a r to line ∴ AP ¯ . 3 i ^ - 2 j ^ + k ^ = 0 3 λ - 5 . 3 + - 2 λ + 2 - 2 + λ - 1 . 1 = 0 9 λ - 15 + 4 λ - 4 + λ - 1 = 0 14 λ = 20 λ = 10 7 ⇒ P 16 7 , 8 7 , 10 7 ⇒ α + β + γ = 16 + 8 + 10 7 = 34 7 21 α + β + γ = 102

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