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The reflection of the point A (1, 0, 0) in the line x12=y+13=z+108 is

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a
(3, –4, –2)
b
(5, –8, –4)
c
(1, –1, –10)
d
(2, –3, 8)

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detailed solution

Correct option is B

Any point P on the line is (2r+1, –3r–1, 8r–10)D.r’s of AP are (2r, –3r–1, 8r–10)AP is perpendicular to the given line2(2r) – 3(–3r–1) + 8(8r–10) = 0     r = 1P (3, –4, –2)Let B be the image of A  B = 2P – A

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