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07/02/2022 12:25 pm
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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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07/02/2022 12:28 pm
\(\frac{x-1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z+1}{6}\) = 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