The shortest distance between the lines

${{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}}$ and

${{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}}$ is :

$$ \begin{aligned} & \mathrm{L}_1: \frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3} \\\\ & \overrightarrow{a_1}=5 \hat{i}+2 \hat{j}+4 \hat{k} \\\\ & \overrightarrow{r_1}=\hat{i}+2 \hat{j}-3 \hat{k} \\\\ & \mathrm{~L}_2: \frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5} \\\\ & \overrightarrow{a_2}=-3 \hat{i}-5 \hat{j}+\hat{k} \\\\ & \overrightarrow{r_2}=\hat{i}+4 \hat{j}-5 \hat{k} \\\\ & \overrightarrow{r_1} \times \overrightarrow{r_2}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{array}\right| \\\\ & =2 \hat{i}+2 \hat{j}+2 \hat{k} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { Shortest distance (d) }=\left|\frac{\left(\overrightarrow{r_1} \times \overrightarrow{r_2}\right) \cdot\left(\overrightarrow{a_1}-\overrightarrow{a_2}\right)}{\left|\overrightarrow{r_1} \times \overrightarrow{r_2}\right|}\right| \\\\ & =\frac{36}{2 \sqrt{3}}=6 \sqrt{3} \end{aligned} $$