The shortest distance between the lines (x-4)/(1) = (y-3)/(2) =...

The given lines are $L_1: \dfrac{x-4}{1} = \dfrac{y-3}{2} = \dfrac{z-2}{-3}$ and $L_2: \dfrac{x+2}{2} = \dfrac{y-6}{4} = \dfrac{z-5}{-5}$.

For $L_1$, a point on the line is $\vec{a}_1 = 4\hat{i} + 3\hat{j} + 2\hat{k}$ and its direction vector is $\vec{b}_1 = \hat{i} + 2\hat{j} - 3\hat{k}$.

For $L_2$, a point on the line is $\vec{a}_2 = -2\hat{i} + 6\hat{j} + 5\hat{k}$ and its direction vector is $\vec{b}_2 = 2\hat{i} + 4\hat{j} - 5\hat{k}$.

The shortest distance $d$ between two skew lines is given by $d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.

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

The cross product of the direction vectors is:

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10 + 12) - \hat{j}(-5 + 6) + \hat{k}(4 - 4) = 2\hat{i} - \hat{j}$.

The magnitude of the cross product is $|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}$.

The dot product is $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (-6)(2) + (3)(-1) + (3)(0) = -12 - 3 = -15$.

Therefore, the shortest distance is $d = \dfrac{|-15|}{\sqrt{5}} = \dfrac{15}{\sqrt{5}} = 3\sqrt{5}$.

Answer: $3\sqrt{5}$