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in Three-dimensional geometry by (36.3k points)

A plane π is perpendicular to the two planes 2x – 2y + z = 0 and x – y + 2z = 4 and passes through the point (1, –2, 1). The distance of π from the point (1, 2, 2) is

(a) 0

(b) 1

(c) √2

(d) 2√2

1 Answer

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Best answer

Answer is (d) 2√2

The equation of any plane passing through (1, – 2, 1) is

a(x – 1) + b (y + 2) + c (z – 1) = 0 ...(i)

The plane (i) is perpendicular to the planes

2x – 2y + z = 0 and x – y + 2z = 4

So, 2a – 2b + c = 0

and a – a + 2c = 0

Thus, 

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