The shortest distance between the lines $$\dfrac{x+7}{-6} = \dfrac{y-6}{7} = z$$ and $$\dfrac{7-x}{2} = y - 2 = z - 6$$ is

Find the shortest distance between the lines:

$$L_1: \frac{x+7}{-6} = \frac{y-6}{7} = z \quad \text{and} \quad L_2: \frac{7-x}{2} = y-2 = z-6$$

For $$L_1$$, the point $$A_1 = (-7, 6, 0)$$ and the direction vector $$\vec{d}_1 = (-6, 7, 1)$$.

For $$L_2$$, rewriting gives $$\frac{x-7}{-2} = \frac{y-2}{1} = \frac{z-6}{1}$$ so the point is $$A_2 = (7, 2, 6)$$ and the direction vector is $$\vec{d}_2 = (-2, 1, 1)$$.

The cross product of the direction vectors is

$$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -6 & 7 & 1 \\ -2 & 1 & 1 \end{vmatrix} = (7-1)\hat{i} - (-6+2)\hat{j} + (-6+14)\hat{k} = (6, 4, 8)$$

Its magnitude is

$$|\vec{d}_1 \times \vec{d}_2| = \sqrt{36 + 16 + 64} = \sqrt{116} = 2\sqrt{29}$$

Next, the vector between the two points is

$$\vec{A_2 - A_1} = (14, -4, 6)$$

The shortest distance is given by

$$d = \frac{|(\vec{A_2 - A_1}) \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|}$$ $$= \frac{|14 \times 6 + (-4) \times 4 + 6 \times 8|}{2\sqrt{29}} = \frac{|84 - 16 + 48|}{2\sqrt{29}} = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}$$

The answer is Option A: $$2\sqrt{29}$$.