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

The shortest distance between two skew lines is given by $$d = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$$

Line 1: $$\vec{a_1} = 5\hat{i} + 2\hat{j} + 4\hat{k}$$, $$\vec{b_1} = \hat{i} + 2\hat{j} - 3\hat{k}$$
Line 2: $$\vec{a_2} = -3\hat{i} - 5\hat{j} + \hat{k}$$, $$\vec{b_2} = \hat{i} + 4\hat{j} - 5\hat{k}$$

$$\vec{a_2} - \vec{a_1} = (-3 - 5)\hat{i} + (-5 - 2)\hat{j} + (1 - 4)\hat{k}$$

$$\vec{a_2} - \vec{a_1} = -8\hat{i} - 7\hat{j} - 3\hat{k}$$

$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix}$$

$$\vec{b_1} \times \vec{b_2} = \hat{i}(-10 - (-12)) - \hat{j}(-5 - (-3)) + \hat{k}(4 - 2)$$

$$\vec{b_1} \times \vec{b_2} = 2\hat{i} + 2\hat{j} + 2\hat{k}$$

$$|\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}$$

$$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-8)(2) + (-7)(2) + (-3)(2)$$

$$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = -16 - 14 - 6 = -36$$

$$d = \left| \frac{-36}{2\sqrt{3}} \right|$$

$$d = 6\sqrt{3}$$