The shortest distance between the lines $$x + 1 = 2y = -12z$$ and $$x = y + 2 = 6z - 6$$ is

Given lines: $$L_1: x + 1 = 2y = -12z$$ and $$L_2: x = y + 2 = 6z - 6$$.

Express lines in symmetric form.

$$L_1$$: Let $$x + 1 = 2y = -12z = t$$. Then $$x = t - 1$$, $$y = \frac{t}{2}$$, $$z = -\frac{t}{12}$$.

Point: $$(-1, 0, 0)$$, Direction: $$(1, \frac{1}{2}, -\frac{1}{12})$$ or $$(12, 6, -1)$$.

$$L_2$$: Let $$x = y + 2 = 6z - 6 = s$$. Then $$x = s$$, $$y = s - 2$$, $$z = \frac{s + 6}{6}$$.

Point: $$(0, -2, 1)$$, Direction: $$(1, 1, \frac{1}{6})$$ or $$(6, 6, 1)$$.

Compute $$\vec{b_1} \times \vec{b_2}$$.

$$\vec{b_1} \times \vec{b_2} = (12, 6, -1) \times (6, 6, 1) = (6+6, -6-12, 72-36) = (12, -18, 36)$$

$$|\vec{b_1} \times \vec{b_2}| = \sqrt{144 + 324 + 1296} = \sqrt{1764} = 42$$

Compute $$\vec{a_2} - \vec{a_1}$$.

$$\vec{a_2} - \vec{a_1} = (0 - (-1), -2 - 0, 1 - 0) = (1, -2, 1)$$

Shortest distance formula.

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

$$= \frac{|1(12) + (-2)(-18) + 1(36)|}{42} = \frac{|12 + 36 + 36|}{42} = \frac{84}{42} = 2$$

Therefore, the correct answer is Option A: $$\mathbf{2}$$.