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
- A14 x+2 y-15 z=1
- B14 x-2 y+15 z=27
- C14 x+2 y+15 z=31
- 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