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