JEE Math Practice Question
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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