The shortest distance between the lines\(\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-2}{-3}\) and \(\frac{x+2}{2}=\frac{y-6}{4}=…

Shortest distance \(=\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)\right|}{\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|}\)

\(\overrightarrow{b_1} \times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|\)

\(=\hat{i}(-10+12)-\hat{j}(-5+6)+\hat{k}(4-4)\)

\(=2 \hat{i}-\hat{j}+0 \hat{k}\)

\(\left|\overrightarrow{b_1} \times \overrightarrow{b_2}\right|=\sqrt{2^2+(-1)^2+0^2}=\sqrt{5}\)

\(\overrightarrow{a_2}-\overrightarrow{a_1}=(-2-4) \hat{i}+(6-3) \hat{j}+(5-2) \hat{k}=-6 \hat{i}+3 \hat{j}+3 \hat{k}\)

\(\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \cdot\left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right)\)

\(=(-6 \hat{i}+3 \hat{j}+3 \hat{k})\cdot (2 \hat{i}-\hat{j}+0 \hat{k})=-15\)

Hence shortest distance

\(d=\frac{15}{\sqrt{5}}=3 \sqrt{5}\)