Let A = (0,0,0), B = (9,8,12), and C = (6,2,3). Find coordinates for the point on line AB that is closest to C.

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  1. Let the perpendicular distance from C to AB meet AB at D
    vector AB = [9,8,12]
    equation of plane through C and D having AB as a normal:
    9x + 8y + 12z = c
    but (6,2,3) lies on it , so the plane is
    9(6)+8(2)+12(3) = c = 106

    plane equation: 9x+8y+12z = 106

    equation of line AB:
    x = 0+9t
    y= 0+8t
    z= 0+12t
    where does AB cut the plane?
    9(9t) + 8(8t) + 12(12t) = 106
    289t = 106
    t = 106/289
    so point D is (954/289 , 848/289 , 1272/289)

    check my arithmetic

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  2. GOOD GRIEF, Just realized I did this same question a week ago.
    At least I ended up with the same answer.

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