Let P be the foot of the perpendicular from the point (1,2,2) on the line L: $$\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}.$$ Let the line $$\vec{r}=(-\hat{i}+\hat{j}-2\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}), \quad \lambda \in \mathbb{R},$$ intersect the line L at Q. Then $$2(PQ)^{2}$$ is equal to:

Line L: $$\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z-2}{2} = t$$

Parametric form: $$(1+t, -1-t, 2+2t)$$

Finding P (foot of perpendicular from (1,2,2) to L):

Direction of L: $$\vec{d} = (1,-1,2)$$

Vector from point on L to (1,2,2): $$(1-(1+t), 2-(-1-t), 2-(2+2t)) = (-t, 3+t, -2t)$$

For perpendicularity: $$(-t)(1) + (3+t)(-1) + (-2t)(2) = 0$$

$$-t - 3 - t - 4t = 0$$

$$-6t - 3 = 0$$

$$t = -\frac{1}{2}$$

So $$P = \left(\frac{1}{2}, -\frac{1}{2}, 1\right)$$.

Finding Q (intersection of given line with L):

The second line: $$\vec{r} = (-1,1,-2) + \lambda(1,-1,1)$$, i.e., $$(-1+\lambda, 1-\lambda, -2+\lambda)$$

Setting equal to L's parametric form:

$$-1+\lambda = 1+t \Rightarrow \lambda = 2+t \quad ...(1)$$

$$1-\lambda = -1-t \Rightarrow \lambda = 2+t \quad ...(2)$$ ✓

$$-2+\lambda = 2+2t \quad ...(3)$$

From (1): $$\lambda = 2+t$$. Substituting in (3):

$$-2+2+t = 2+2t$$

$$t = -2$$

$$\lambda = 0$$

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

Verification on L: $$(1+(-2), -1-(-2), 2+2(-2)) = (-1, 1, -2)$$ ✓

Computing $$2(PQ)^2$$:

$$PQ = Q - P = \left(-1-\frac{1}{2}, 1+\frac{1}{2}, -2-1\right) = \left(-\frac{3}{2}, \frac{3}{2}, -3\right)$$

$$(PQ)^2 = \frac{9}{4} + \frac{9}{4} + 9 = \frac{9}{4} + \frac{9}{4} + \frac{36}{4} = \frac{54}{4} = \frac{27}{2}$$

$$2(PQ)^2 = 2 \times \frac{27}{2} = 27$$

The correct answer is Option 4: 27.