The distance of the point (1, 1, 9) from the point of intersection of the line $$\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$$ and the plane $$x + y + z = 17$$ is:

We are given the line $$\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2} = \lambda$$ and the plane $$x + y + z = 17$$. We need to find the distance from $$(1, 1, 9)$$ to their point of intersection.

Any point on the line can be written as $$(3 + \lambda, \, 4 + 2\lambda, \, 5 + 2\lambda)$$.

Substituting this into the plane equation, we get $$(3 + \lambda) + (4 + 2\lambda) + (5 + 2\lambda) = 17$$.

Simplifying, $$12 + 5\lambda = 17$$, so $$\lambda = 1$$.

The point of intersection is $$(3 + 1, \, 4 + 2, \, 5 + 2) = (4, 6, 7)$$.

Now we find the distance between $$(1, 1, 9)$$ and $$(4, 6, 7)$$. Using the distance formula, $$d = \sqrt{(4 - 1)^2 + (6 - 1)^2 + (7 - 9)^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$$.

Hence, the correct answer is Option C.