If the shortest distance between the lines $$\frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1}$$ and $$\frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1}$$ is $$1$$, then the sum of all possible values of $$\lambda$$ is:

The shortest distance between two skew lines is given by the formula:

$$d = \left| \frac{ (\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) }{ |\vec{b_1} \times \vec{b_2}| } \right|$$

where $$\vec{a_1}$$ and $$\vec{a_2}$$ are points on the lines, and $$\vec{b_1}$$ and $$\vec{b_2}$$ are the direction vectors.

For the first line: $$\frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1}$$, a point is $$\vec{a_1} = \langle \lambda, 2, 1 \rangle$$ and direction vector $$\vec{b_1} = \langle -2, 1, 1 \rangle$$.

For the second line: $$\frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1}$$, a point is $$\vec{a_2} = \langle \sqrt{3}, 1, 2 \rangle$$ and direction vector $$\vec{b_2} = \langle 1, -2, 1 \rangle$$.

The vector between the points is $$\vec{a_2} - \vec{a_1} = \langle \sqrt{3} - \lambda, -1, 1 \rangle$$.

Compute the cross product $$\vec{b_1} \times \vec{b_2}$$:

$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - 1 \cdot (-2)) - \hat{j}(-2 \cdot 1 - 1 \cdot 1) + \hat{k}(-2 \cdot (-2) - 1 \cdot 1)$$

$$= \hat{i}(1 + 2) - \hat{j}(-2 - 1) + \hat{k}(4 - 1) = \hat{i}(3) - \hat{j}(-3) + \hat{k}(3) = \langle 3, 3, 3 \rangle$$

The magnitude is $$|\vec{b_1} \times \vec{b_2}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$$.

The scalar triple product is:

$$(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) = \langle 3, 3, 3 \rangle \cdot \langle \sqrt{3} - \lambda, -1, 1 \rangle = 3(\sqrt{3} - \lambda) + 3(-1) + 3(1) = 3\sqrt{3} - 3\lambda - 3 + 3 = 3\sqrt{3} - 3\lambda$$

The distance is:

$$d = \left| \frac{3\sqrt{3} - 3\lambda}{3\sqrt{3}} \right| = \left| \frac{\sqrt{3} - \lambda}{\sqrt{3}} \right| = \left| 1 - \frac{\lambda}{\sqrt{3}} \right|$$

Given that $$d = 1$$:

$$\left| 1 - \frac{\lambda}{\sqrt{3}} \right| = 1$$

This absolute value equation gives two cases:

Case 1: $$1 - \frac{\lambda}{\sqrt{3}} = 1$$

$$-\frac{\lambda}{\sqrt{3}} = 0 \implies \lambda = 0$$

Case 2: $$1 - \frac{\lambda}{\sqrt{3}} = -1$$

$$-\frac{\lambda}{\sqrt{3}} = -2 \implies \frac{\lambda}{\sqrt{3}} = 2 \implies \lambda = 2\sqrt{3}$$

The possible values of $$\lambda$$ are $$0$$ and $$2\sqrt{3}$$. The sum is $$0 + 2\sqrt{3} = 2\sqrt{3}$$.

Verification:

For $$\lambda = 0$$, the numerator is $$3\sqrt{3} - 3(0) = 3\sqrt{3}$$, so $$d = \left| \frac{3\sqrt{3}}{3\sqrt{3}} \right| = 1$$.

For $$\lambda = 2\sqrt{3}$$, the numerator is $$3\sqrt{3} - 3(2\sqrt{3}) = 3\sqrt{3} - 6\sqrt{3} = -3\sqrt{3}$$, so $$d = \left| \frac{-3\sqrt{3}}{3\sqrt{3}} \right| = 1$$.

Thus, the sum of all possible values of $$\lambda$$ is $$2\sqrt{3}$$.