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The square of the distance of the point of intersection of the line x-1/2 = y-2/3 = z+1/6 and the plane 2x – y + z = 6 from the point (–1, –1, 2) is

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The square of the distance of the point of intersection of the line \(\frac{x-1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z+1}{6}\) and the plane 2x – y + z = 6 from the point (–1, –1, 2) is _________.

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\(\frac{x-1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z+1}{6}\) = λ

x = 2λ + 1, y = 3λ, z = 6λ - 1

for point of intersection of line & plane

2(2λ + 1) – (3λ + 2) + (6λ – 1) = 6

7λ = 7 ⇒ λ = 1

point : (3, 5, 5)

(distance)2 = (3 + 1)2 + (5 + 1)2 + (5 – 2)2

= 16 + 36 + 9 = 61

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