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MHT CET20249 May 2024Morning ShiftMathematicsThree Dimensional GeometryActual

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

Options

  1. A14 x+2 y-15 z=1
  2. B14 x-2 y+15 z=27
  3. C14 x+2 y+15 z=31
  4. D-14 x+2 y+15 z=3

Correct answer

C. 14 x+2 y+15 z=31

Step-by-step solution

The equation of plane passing through (1,1,1) is a(x-1)+b(y-1)+c(z-1)=0 Since plane (i) is perpendicular to the planes 2 x+y-2 z=5 and 3 x-6 y-2 z=7 2 a + b -2 c =5 ...(i) 3 a-6 b-2 c=7...(ii) On solving (i), (ii) and (iii), we get a=14, b=2, c=15 Substituting the values of a, b, c in (i), we get aligned & 14(x-1)+2(y-1)+15(z-1)=0 & 14 x+2 y+15 z=31 aligned

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