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 _____

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 $$L_1$$: Point $$\vec{a_1} = (2, -1, 6)$$, direction $$\vec{b_1} = (3, 2, 2)$$.

For $$L_2$$: Rewriting $$\frac{1-y}{2} = \frac{y-1}{-2}$$, so direction $$\vec{b_2} = (3, -2, 0)$$. Point $$\vec{a_2} = (6, 1, -8)$$.

The formula for the shortest distance between two skew lines is:

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

$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{vmatrix}$$

$$= \hat{i}(2 \cdot 0 - 2 \cdot (-2)) - \hat{j}(3 \cdot 0 - 2 \cdot 3) + \hat{k}(3 \cdot (-2) - 2 \cdot 3)$$

$$= \hat{i}(0 + 4) - \hat{j}(0 - 6) + \hat{k}(-6 - 6) = (4, 6, -12)$$

$$|\vec{b_1} \times \vec{b_2}| = \sqrt{4^2 + 6^2 + (-12)^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$$

$$\vec{a_2} - \vec{a_1} = (6-2, 1-(-1), -8-6) = (4, 2, -14)$$

$$(\vec{a_2} - \vec{a_1}) \cdot (4, 6, -12) = 4(4) + 2(6) + (-14)(-12) = 16 + 12 + 168 = 196$$

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

The shortest distance is 14.