The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x+y-2z=5 and 3x-6y-2z=7, is

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−14x+2y+15z=3

14x−2y+15z=27

14x+2y−15z=1

14x+2y+15z=31

answer is D.

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Detailed Solution

ax−1+by−1+cz−1=02a+b−2c=0,3a−6b−2c=0a−14=+b−2=c−15a:b:c=14:2:1514x+2y+15z=14+2+15=31

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