MathematicsMedium249×since 2002Q2598The foot of the perpendicular from a point on the circle x2+y2=1,z=0x^{2}+y^{2}=1, z=0x2+y2=1,z=0 to the plane 2x+3y+z=62 x+3 y+z=62x+3y+z=6 lies on which one of the following curves?A(6x+5y−12)2+4(3x+7y−8)2=1,z=6−2x−3y(6 x+5 y-12)^{2}+4(3 x+7 y-8)^{2}=1, z=6-2 x-3 y(6x+5y−12)2+4(3x+7y−8)2=1,z=6−2x−3yB(5x+6y−12)2+4(3x+5y−9)2=1,z=6−2x−3y(5 x+6 y-12)^{2}+4(3 x+5 y-9)^{2}=1, z=6-2 x-3 y(5x+6y−12)2+4(3x+5y−9)2=1,z=6−2x−3yC(6x+5y−14)2+9(3x+5y−7)2=1,z=6−2x−3y(6 x+5 y-14)^{2}+9(3 x+5 y-7)^{2}=1, z=6-2 x-3 y(6x+5y−14)2+9(3x+5y−7)2=1,z=6−2x−3yD(5x+6y−14)2+9(3x+7y−8)2=1,z=6−2x−3y(5 x+6 y-14)^{2}+9(3 x+7 y-8)^{2}=1, z=6-2 x-3 y(5x+6y−14)2+9(3x+7y−8)2=1,z=6−2x−3yCheck answerSkip