Let the line L intersect the lines $$x - 2 = -y = z - 1$$, $$2(x + 1) = 2(y - 1) = z + 1$$ and be parallel to the line $$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{2}$$. Then which of the following points lies on L?

We are given that line L intersects $$L_1: x-2 = -y = z-1$$ and $$L_2: 2(x+1) = 2(y-1) = z+1$$, and is parallel to $$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{2}$$. Our goal is to determine which point lies on L.

First, note that $$L_1$$ can be written in symmetric form as $$\frac{x-2}{1} = \frac{y}{-1} = \frac{z-1}{1}$$, revealing it passes through $$(2,0,1)$$ with direction vector $$(1,-1,1)$$, while $$L_2$$ is given by $$\frac{x+1}{1} = \frac{y-1}{1} = \frac{z+1}{2}$$, passing through $$(-1,1,-1)$$ with direction $$(1,1,2)$$. Since L is parallel to $$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z-2}{2}$$, its direction vector is $$(3,1,2)$$.

Next, we let a general point on $$L_1$$ be $$(2+t,-t,1+t)$$ and a general point on $$L_2$$ be $$(-1+s,1+s,-1+2s)$$. The vector joining these points must be proportional to $$(3,1,2)$$, which leads to

$$\frac{-3+s-t}{3} = \frac{1+s+t}{1} = \frac{-2+2s-t}{2}.$$

Equating the middle and last terms, $$\frac{1+s+t}{1} = \frac{-2+2s-t}{2}$$ becomes $$2(1+s+t) = -2+2s-t$$, so $$3t = -4$$ and $$t = -\frac{4}{3}$$.

Then equating the first and middle terms, $$\frac{-3+s-t}{3} = \frac{1+s+t}{1}$$ gives $$-3+s-t = 3+3s+3t$$, which simplifies to $$-6 = 2s+4t$$ and thus $$s = -\frac{1}{3}$$.

Substituting $$t = -\frac{4}{3}$$ into the point on $$L_1$$ yields $$\left(2 - \frac{4}{3},\frac{4}{3},1 - \frac{4}{3}\right) = \left(\frac{2}{3},\frac{4}{3},-\frac{1}{3}\right).$$

Since L passes through $$\left(\frac{2}{3},\frac{4}{3},-\frac{1}{3}\right)$$ with direction $$(3,1,2)$$, we test Option (1): $$\left(-\frac{1}{3},1,-1\right)$$. The vector from our point to this candidate is $$\left(-\frac{1}{3}-\frac{2}{3},\,1-\frac{4}{3},\,-1+\frac{1}{3}\right) = \left(-1,-\frac{1}{3},-\frac{2}{3}\right), $$ which is proportional to $$(3,1,2)$$ via the factor $$-\frac{1}{3}$$, confirming that this point lies on L.

The correct answer is Option (1): $$\left(-\frac{1}{3}, 1, -1\right)$$.