The foot of the perpendicular from a point on the circle \( x^2 + y^2 = 1, \, z = 0 \) to the plane \( 2x + 3y + z = 6 \) lies on which one of the following curves?
\( (6x+5y-12)^2 + 4(3x+7y-8)^2 = 1, \; z = 6-2x-3y \)
\( (5x+6y-12)^2 + 4(3x+5y-9)^2 = 1, \; z = 6-2x-3y \)
\( (6x+5y-14)^2 + 9(3x+5y-7)^2 = 1, \; z = 6-2x-3y \)
\( (5x+6y-14)^2 + 9(3x+7y-8)^2 = 1, \; z = 6-2x-3y \)