The reflection of the point A (1, 0, 0) in the line x−12=y+1−3=z+108 is

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