1.Foot of perpendicular from a point on x2+y2=1,z=0x^2+y^2=1,z=0x2+y2=1,z=0 to plane 2x+3y+z=62x+3y+z=62x+3y+z=6 lies on which curve?a.(6x+5y−12)2+4(3x+7y−8)2=1,z=6−2x−3y(6x+5y-12)^2+4(3x+7y-8)^2=1,z=6-2x-3y(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(5x+6y-12)^2+4(3x+5y-9)^2=1,z=6-2x-3y(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(6x+5y-14)^2+9(3x+5y-7)^2=1,z=6-2x-3y(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(5x+6y-14)^2+9(3x+7y-8)^2=1,z=6-2x-3y(5x+6y−14)2+9(3x+7y−8)2=1,z=6−2x−3yLogin to continueOnly logged in users canattempt or see the solution.
