The shortest distance between the lines $$\frac{x+2}{1} = \frac{y}{-2} = \frac{z-5}{2}$$ and $$\frac{x-4}{1} = \frac{y-1}{2} = \frac{z+3}{0}$$ is

Line 1: point $$(-2,0,5)$$, direction $$(1,-2,2)$$

Line 2: point $$(4,1,-3)$$, direction $$(1,2,0)$$

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

$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 1 & 2 & 0 \end{vmatrix} = (0-4, 2-0, 2+2) = (-4, 2, 4)$$

$$|\vec{b_1} \times \vec{b_2}| = \sqrt{16 + 4 + 16} = 6$$

Shortest distance:

$$d = \frac{|(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} = \frac{|6(-4) + 1(2) + (-8)(4)|}{6} = \frac{|-24 + 2 - 32|}{6} = \frac{54}{6} = 9$$

The correct answer is Option 4: 9.