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?

<p>Any point on ${x^2} + {y^2} = 1$, $z = 0$ is $p(\cos \theta ,\,\sin \theta ,\,0)$</p> <p>If foot of perpendicular of p on the plane $2x + 3y + z = 6$ is $(h,k,l)$ then</p> <p>${{h - \cos \theta } \over 2} = {{k - \sin \theta } \over 3} = {{l - 0} \over 1}$</p> <p>$$ = - \left( {{{2\cos \theta + 3\sin \theta + 0 - 6} \over {{2^2} + {3^2} + {1^2}}}} \right) = r$$ (let)</p> <p>$h = 2r + \cos \theta ,\,k = 3r + \sin \theta ,\,l = r$</p> <p>Hence, $h - 2l = \cos \theta$ and $k - 3l = \sin \theta$</p> <p>Hence, ${(h - 2l)^2} + {(k - 3l)^2} = 1$</p> <p>When $l = 6 - 2h - 3k$</p> <p>Hence required locus is</p> <p>${(x - 2(6 - 2x - 3y))^2} + {(y - 3(6 - 2x - 3y))^2} = 1$</p> <p>$\Rightarrow {(5x + 6y - 12)^2} + 4{(3x + 5y - 9)^2} = 1,\,z = 6 - 2x - 3y$</p>