The shortest distance between the lines $$\dfrac{x - 4}{1} = \dfrac{y - 3}{2} = \dfrac{z - 2}{-3}$$ and $$\dfrac{x + 2}{2} = \dfrac{y - 6}{4} = \dfrac{z - 5}{-5}$$ is :

The shortest distance ($$d$$) between two skew lines $$\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$$ and $$\vec{r} = \vec{a}_2 + \mu \vec{b}_2$$ is given by the formula:

$$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$

From the first line $$\frac{x - 4}{1} = \frac{y - 3}{2} = \frac{z - 2}{-3}$$:

$$\vec{a}_1 = 4\hat{i} + 3\hat{j} + 2\hat{k}$$

$$\vec{b}_1 = 1\hat{i} + 2\hat{j} - 3\hat{k}$$

$$\vec{a}_2 = -2\hat{i} + 6\hat{j} + 5\hat{k}$$

$$\vec{b}_2 = 2\hat{i} + 4\hat{j} - 5\hat{k}$$

$$\vec{a}_2 - \vec{a}_1 = (-2 - 4)\hat{i} + (6 - 3)\hat{j} + (5 - 2)\hat{k} = -6\hat{i} + 3\hat{j} + 3\hat{k}$$

$$\vec{b}_1 \times \vec{b}_2 = \text{det} \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{bmatrix}$$

$$\vec{b}_1 \times \vec{b}_2 = 2\hat{i} - 1\hat{j} + 0\hat{k}$$

$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1} = \sqrt{5}$$

$$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-6)(2) + (3)(-1) + (3)(0)$$

$$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = -12 - 3 + 0 = -15$$

$$d = \frac{|-15|}{\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$$