The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x+y−2z=5and3x−6y−2z=7 is
A
14x+2y−15z=1
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B
14x−2y+15z=27
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C
14x+2y+15z=31
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D
−14x+2y+15z=3
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Solution
The correct option is C14x+2y+15z=31 Let the required plane be a(x–1)+b(y–1)+c(z–1)=0
Direction ratio of its normal is ∣∣
∣
∣∣^i^j^k21−23−6−2∣∣
∣
∣∣=−14^i−2^j−15^k
∴ Equation of the plane is −14(x–1)−2(y–1)−15(z–1)=0⇒14x+2y+15z=31