The shortest distance between the lines $$\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$$ and $$\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$$, lies in the interval:

The shortest distance between two skew lines can be found using the formula:

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

First, express the given lines in vector form. The first line is $$\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$$, which can be rewritten as $$\frac{x - 0}{2} = \frac{y - 0}{2} = \frac{z - 0}{1}$$. So, a point on this line is $$\vec{a_1} = (0, 0, 0)$$ and the direction vector is $$\vec{b_1} = (2, 2, 1)$$.

The second line is $$\frac{x + 2}{-1} = \frac{y - 4}{8} = \frac{z - 5}{4}$$, which can be rewritten as $$\frac{x - (-2)}{-1} = \frac{y - 4}{8} = \frac{z - 5}{4}$$. So, a point on this line is $$\vec{a_2} = (-2, 4, 5)$$ and the direction vector is $$\vec{b_2} = (-1, 8, 4)$$.

Now, compute $$\vec{a_2} - \vec{a_1}$$:

$$ \vec{a_2} - \vec{a_1} = (-2 - 0, 4 - 0, 5 - 0) = (-2, 4, 5) $$

Next, compute the cross product $$\vec{b_1} \times \vec{b_2}$$. Using the determinant formula:

$$ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ -1 & 8 & 4 \end{vmatrix} $$

Expand the determinant:

$$ \hat{i} (2 \cdot 4 - 1 \cdot 8) - \hat{j} (2 \cdot 4 - 1 \cdot (-1)) + \hat{k} (2 \cdot 8 - 2 \cdot (-1)) $$

Calculate each component:

For $$\hat{i}$$: $$2 \cdot 4 - 1 \cdot 8 = 8 - 8 = 0$$

For $$\hat{j}$$: $$-(2 \cdot 4 - 1 \cdot (-1)) = -(8 - (-1)) = -(8 + 1) = -9$$

For $$\hat{k}$$: $$2 \cdot 8 - 2 \cdot (-1) = 16 - (-2) = 16 + 2 = 18$$

So, $$\vec{b_1} \times \vec{b_2} = (0, -9, 18)$$

Now, compute the magnitude of this cross product:

$$ | \vec{b_1} \times \vec{b_2} | = \sqrt{0^2 + (-9)^2 + 18^2} = \sqrt{0 + 81 + 324} = \sqrt{405} $$

Simplify $$\sqrt{405}$$:

$$ \sqrt{405} = \sqrt{81 \times 5} = \sqrt{81} \times \sqrt{5} = 9\sqrt{5} $$

Next, compute the dot product $$(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})$$:

$$ (0, -9, 18) \cdot (-2, 4, 5) = 0 \cdot (-2) + (-9) \cdot 4 + 18 \cdot 5 = 0 - 36 + 90 = 54 $$

The absolute value is $$|54| = 54$$.

Now, substitute into the distance formula:

$$ d = \frac{54}{9\sqrt{5}} = \frac{6}{\sqrt{5}} $$

Rationalize the denominator:

$$ d = \frac{6}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{6\sqrt{5}}{5} $$

So, the shortest distance is $$\frac{6\sqrt{5}}{5}$$.

To find the interval, compute the numerical value. $$\sqrt{5} \approx 2.236$$:

$$ d = \frac{6 \times 2.236}{5} = \frac{13.416}{5} = 2.6832 $$

Check the intervals:

Option A: $$(3, 4]$$ → 2.6832 is not in this interval.

Option B: $$(2, 3]$$ → 2.6832 is greater than 2 and less than or equal to 3, so it lies in this interval.

Option C: $$[1, 2)$$ → 2.6832 is greater than 2, so not in this interval.

Option D: $$[0, 1)$$ → 2.6832 is greater than 1, so not in this interval.

Hence, the shortest distance lies in the interval $$(2, 3]$$.

So, the answer is Option B.