Equation of the line of the shortest distance between the lines $$\frac{x}{1} = \frac{y}{-1} = \frac{z}{1}$$ and $$\frac{x-1}{0} = \frac{y+1}{-2} = \frac{z}{1}$$ is:

To find the equation of the line of the shortest distance between the two given lines, we first identify points and direction vectors for each line.

The first line is given as $$\frac{x}{1} = \frac{y}{-1} = \frac{z}{1}$$. This can be rewritten as $$\frac{x-0}{1} = \frac{y-0}{-1} = \frac{z-0}{1}$$. So, a point on this line is $$A(0, 0, 0)$$, and the direction vector is $$\vec{d_1} = (1, -1, 1)$$.

The second line is $$\frac{x-1}{0} = \frac{y+1}{-2} = \frac{z}{1}$$. Since the denominator for $$x$$ is 0, the $$x$$-coordinate is constant at 1. Letting the parameter be $$\lambda$$, we have $$x = 1$$, $$y + 1 = -2\lambda$$ so $$y = -1 - 2\lambda$$, and $$z = \lambda$$. Thus, a point on this line is $$B(1, -1, 0)$$, and the direction vector is $$\vec{d_2} = (0, -2, 1)$$.

The line of shortest distance is perpendicular to both given lines. Therefore, its direction vector $$\vec{d}$$ is the cross product of $$\vec{d_1}$$ and $$\vec{d_2}$$.

Compute $$\vec{d} = \vec{d_1} \times \vec{d_2}$$:

$$$ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 0 & -2 & 1 \\ \end{vmatrix} $$$

Expanding the determinant:

$$$ \hat{i} \left[ (-1)(1) - (1)(-2) \right] - \hat{j} \left[ (1)(1) - (1)(0) \right] + \hat{k} \left[ (1)(-2) - (-1)(0) \right] $$$

$$$ = \hat{i} \left[ -1 - (-2) \right] - \hat{j} \left[ 1 - 0 \right] + \hat{k} \left[ -2 - 0 \right] $$$

$$$ = \hat{i} (1) - \hat{j} (1) + \hat{k} (-2) $$$

$$$ = (1, -1, -2) $$$

So, the direction vector of the line of shortest distance is $$(1, -1, -2)$$.

Next, we find points where this line intersects the given lines. Let a point on the first line be $$P(t) = (t, -t, t)$$, and a point on the second line be $$Q(s) = (1, -1 - 2s, s)$$. The vector $$\overrightarrow{PQ}$$ is:

$$$ \overrightarrow{PQ} = (1 - t, (-1 - 2s) - (-t), s - t) = (1 - t, -1 - 2s + t, s - t) $$$

Since $$\overrightarrow{PQ}$$ is parallel to $$\vec{d} = (1, -1, -2)$$, it must be perpendicular to both $$\vec{d_1}$$ and $$\vec{d_2}$$.

First, $$\overrightarrow{PQ} \cdot \vec{d_1} = 0$$:

$$$ (1 - t)(1) + (-1 - 2s + t)(-1) + (s - t)(1) = 0 $$$

Simplify:

$$$ 1 - t + (1 + 2s - t) + s - t = 0 $$$

$$$ 1 - t + 1 + 2s - t + s - t = 0 $$$

$$$ 2 + 3s - 3t = 0 $$$

$$$ 3s - 3t = -2 $$$

$$$ s - t = -\frac{2}{3} \quad \text{(Equation 1)} $$$

Second, $$\overrightarrow{PQ} \cdot \vec{d_2} = 0$$:

$$$ (1 - t)(0) + (-1 - 2s + t)(-2) + (s - t)(1) = 0 $$$

Simplify:

$$$ 0 + (-2)(-1 - 2s + t) + s - t = 0 $$$

$$$ 2 + 4s - 2t + s - t = 0 $$$

$$$ 2 + 5s - 3t = 0 \quad \text{(Equation 2)} $$$

Solve Equations 1 and 2 simultaneously. From Equation 1:

$$$ t = s + \frac{2}{3} $$$

Substitute into Equation 2:

$$$ 2 + 5s - 3\left(s + \frac{2}{3}\right) = 0 $$$

$$$ 2 + 5s - 3s - 2 = 0 $$$

$$$ 2s = 0 $$$

$$$ s = 0 $$$

Then:

$$$ t = 0 + \frac{2}{3} = \frac{2}{3} $$$

So, point $$P$$ is $$\left(\frac{2}{3}, -\frac{2}{3}, \frac{2}{3}\right)$$ and point $$Q$$ is $$(1, -1, 0)$$.

The line of shortest distance passes through $$Q(1, -1, 0)$$ with direction vector $$(1, -1, -2)$$. Its equation is:

$$$ \frac{x - 1}{1} = \frac{y - (-1)}{-1} = \frac{z - 0}{-2} $$$

$$$ \frac{x - 1}{1} = \frac{y + 1}{-1} = \frac{z}{-2} $$$

Comparing with the options, this matches option C.

Hence, the correct answer is Option C.