The shortest distance between the lines $$\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-6}{2}$$ and $$\frac{x-6}{3} = \frac{1-y}{2} = \frac{z+8}{0}$$ is equal to ______

We need to find the shortest distance between the lines:

$$L_1: \frac{x-2}{3} = \frac{y+1}{2} = \frac{z-6}{2}$$

$$L_2: \frac{x-6}{3} = \frac{1-y}{2} = \frac{z+8}{0}$$

For $$L_1$$, we take the point $$A = (2, -1, 6)$$ and the direction vector $$\vec{b_1} = (3, 2, 2)$$, and for $$L_2$$, we take the point $$B = (6, 1, -8)$$ with the direction vector $$\vec{b_2} = (3, -2, 0)$$.

The cross product of the direction vectors is

$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{vmatrix} = \hat{i}(0+4) - \hat{j}(0-6) + \hat{k}(-6-6) = (4, 6, -12)$$

The vector joining the two chosen points is

$$\vec{AB} = B - A = (4, 2, -14)$$

Applying the formula for the shortest distance between skew lines, we have

$$d = \frac{|\vec{AB} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$$

The dot product in the numerator is

$$\vec{AB} \cdot (4, 6, -12) = 4(4) + 2(6) + (-14)(-12) = 16 + 12 + 168 = 196$$

The magnitude of the cross product in the denominator is

$$|\vec{b_1} \times \vec{b_2}| = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$$

Therefore,

$$d = \frac{|196|}{14} = 14$$

The answer is 14.