Let P and Q be the points on the line x + 3 8 = y − 4 2 = z + 1 2 which are at a distance of 6 units from the point R ( 1 , 2 , 3 ) . If the centroid of the triangle P Q R is α , β , γ , then α 2 + β 2 + γ 2 is:

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Correct answer: C. 18

Let, x + 3 8 = y − 4 2 = z + 1 2 = λ . ⇒ x = 8 λ - 3 , y = 2 λ + 4 , z = 2 λ - 1 Let, P ≡ 8 λ - 3 , 2 λ + 4 , 2 λ - 1 . Now, given R 1 , 2 , 3 and P R = 6 & Q R = 6 . So, by distance formula we get, ⇒ 8 λ - 4 2 + 2 λ + 2 2 + 2 λ - 4 2 = 36 ⇒ 64 λ 2 + 16 - 64 λ + 4 λ 2 + 4 + 8 λ + 4 λ 2 + 16 - 16 λ = 36 ⇒ 72 λ 2 - 72 λ = 0 ⇒ λ = 0 , 1 ⇒ P ≡ - 3 , 4 , - 1 and Q ≡ 5 , 6 , 1 Hence, centroid of △ P Q R will be, α , β , γ ≡ 1 , 4 , 1 . ⇒ α 2 + β 2 + γ 2 = 18

Let P and Q be the points on the line x + 3 8 = y − 4 2 = z + 1 2 which are at a distance of 6 units from the point R ( 1 , 2 , 3 ) . If the centroid of the triangle P Q R is α , β , γ , then α 2 + β 2 + γ 2 is:

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