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?

We need to find the locus of the foot of the perpendicular from a point on the circle $$x^2 + y^2 = 1, z = 0$$ to the plane $$2x + 3y + z = 6$$.

Let $$P = (\cos\theta, \sin\theta, 0)$$ be a point on the circle.

The normal to the plane $$2x + 3y + z = 6$$ is $$\vec{n} = (2, 3, 1)$$. Since $$Q = (X, Y, Z)$$ is the foot of the perpendicular from $$P$$, it satisfies

$$\frac{X - \cos\theta}{2} = \frac{Y - \sin\theta}{3} = \frac{Z - 0}{1} = -\frac{2\cos\theta + 3\sin\theta + 0 - 6}{4 + 9 + 1}$$

$$= \frac{6 - 2\cos\theta - 3\sin\theta}{14}$$

Let $$\lambda = \frac{6 - 2\cos\theta - 3\sin\theta}{14}$$. This gives us

$$X = \cos\theta + 2\lambda, \quad Y = \sin\theta + 3\lambda, \quad Z = \lambda.$$

Since $$Q$$ lies on the plane, we also have $$Z = 6 - 2X - 3Y$$.

From the expressions for $$X$$, $$Y$$, and $$Z$$ it follows that $$\cos\theta = X - 2Z, \quad \sin\theta = Y - 3Z$$.

Using the identity $$\cos^2\theta + \sin^2\theta = 1$$ gives

$$ (X - 2Z)^2 + (Y - 3Z)^2 = 1. $$

Substituting $$Z = 6 - 2X - 3Y$$ leads to

$$ (X - 2(6 - 2X - 3Y))^2 + (Y - 3(6 - 2X - 3Y))^2 = 1, $$

which simplifies to

$$ (X - 12 + 4X + 6Y)^2 + (Y - 18 + 6X + 9Y)^2 = 1, $$

or

$$ (5X + 6Y - 12)^2 + (6X + 10Y - 18)^2 = 1, $$

and hence

$$ (5X + 6Y - 12)^2 + 4(3X + 5Y - 9)^2 = 1 $$

With the constraint $$z = 6 - 2x - 3y$$, this equation represents the required locus, corresponding to Option B.