From the point P(a, b, c),let perpendiculars PL and PM be drawn to YOZ and ZOX planes, respectively. Then theequation of the plane OLM is
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a
xa+yb+zc=0
b
xa+yb−zc=0
c
xa−yb−zc=0
d
xa−yb+zc=0
answer is B.
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Detailed Solution
Coordinates of L and M are (0, b, c) and (a,0, c), respectively. Therefore, the equation of the plane passing through (0, 0, 0), (0, b, c) and (a, 0, c) is x−0y−0z−00bca0c=0 or xa+yb−zc=0