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Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). The image of R in the plane P is:

1. (6, 5, 2)
2. (6, 5, −2)
3. (4, 3, 2)
4. (3, 4, −2)

Correct answer: (6, 5, −2)

Solution

Normal = (B-A)x(C-A) = (1,1,-2), so plane x+y-2z-3=0. For R=(2,1,6): plane value = 2+1-12-3 = -12, |n|^2=6, t=-12/6=-2; image = R-2t*n = (2,1,6)+4*(1,1,-2) = (6,5,-2).

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