Let Q(a, b, c) be the image of the point P(3, 2, 1) in the line $$ \frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}$$ Then the distance of Q from the line $$ \frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2} $$ is

To find the image $$Q(a, b, c)$$ of point $$P(3, 2, 1)$$ in the line $$\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}$$, first determine the parametric equations of the line. Let $$\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1} = \lambda$$. Thus, any point on the line is $$(1 + \lambda, 2\lambda, 1 + \lambda)$$.

The direction vector of the line is $$\vec{d} = (1, 2, 1)$$. Let $$M$$ be the foot of the perpendicular from $$P$$ to the line. The vector $$\overrightarrow{PM}$$ is perpendicular to $$\vec{d}$$, so their dot product is zero.

Vector $$\overrightarrow{PM} = (1 + \lambda - 3, 2\lambda - 2, 1 + \lambda - 1) = (\lambda - 2, 2\lambda - 2, \lambda)$$.
Dot product: $$(1)(\lambda - 2) + (2)(2\lambda - 2) + (1)(\lambda) = 0$$
$$\lambda - 2 + 4\lambda - 4 + \lambda = 0$$
$$6\lambda - 6 = 0$$
$$\lambda = 1$$.

Thus, $$M = (1 + 1, 2 \cdot 1, 1 + 1) = (2, 2, 2)$$.

Since $$M$$ is the midpoint of $$P$$ and $$Q$$,
$$\frac{3 + a}{2} = 2 \implies 3 + a = 4 \implies a = 1$$
$$\frac{2 + b}{2} = 2 \implies 2 + b = 4 \implies b = 2$$
$$\frac{1 + c}{2} = 2 \implies 1 + c = 4 \implies c = 3$$.

So, $$Q = (1, 2, 3)$$.

Now, find the distance of $$Q(1, 2, 3)$$ from the line $$\frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2}$$.

Let $$A(9, 9, 5)$$ be a point on the line, and $$\vec{d_2} = (3, 2, -2)$$ be its direction vector. The vector $$\overrightarrow{AQ} = (1 - 9, 2 - 9, 3 - 5) = (-8, -7, -2)$$.

The distance from a point to a line is given by $$\frac{\|\overrightarrow{AQ} \times \vec{d_2}\|}{\|\vec{d_2}\|}$$.

Compute the cross product:
$$\overrightarrow{AQ} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -8 & -7 & -2 \\ 3 & 2 & -2 \end{vmatrix}$$
$$= \hat{i}[(-7) \cdot (-2) - (-2) \cdot 2] - \hat{j}[(-8) \cdot (-2) - (-2) \cdot 3] + \hat{k}[(-8) \cdot 2 - (-7) \cdot 3]$$
$$= \hat{i}[14 - (-4)] - \hat{j}[16 - (-6)] + \hat{k}[-16 - (-21)]$$
$$= \hat{i}(18) - \hat{j}(22) + \hat{k}(5)$$
$$= (18, -22, 5)$$.

Magnitude: $$\|(18, -22, 5)\| = \sqrt{18^2 + (-22)^2 + 5^2} = \sqrt{324 + 484 + 25} = \sqrt{833}$$.
Factorize: $$833 = 49 \times 17$$, so $$\sqrt{833} = \sqrt{49 \times 17} = 7\sqrt{17}$$.

Magnitude of $$\vec{d_2}$$: $$\|(3, 2, -2)\| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17}$$.

Distance: $$\frac{7\sqrt{17}}{\sqrt{17}} = 7$$.

The distance is 7, which corresponds to option B.