The shortest distance between  the lines  $\overrightarrow{r}=3\overrightarrow{i}-15\overrightarrow{j}+9\overrightarrow{k}+\lambda \left(2\overrightarrow{i}-7\overrightarrow{j}+5\overrightarrow{k}\right)$ and $\overrightarrow{r}=\left(-\overrightarrow{i}+\overrightarrow{j}+9\overrightarrow{k}\right)+\mu \left(2\overrightarrow{i}+\overrightarrow{j}-3\overrightarrow{k}\right)$ is

$$\begin{gathered} Given lines are \\ \overline{r}=3\overline{i}-15\overline{j}+9\overline{k}+\lambda (2\overline{i}-7\overline{j}+5\overline{k}) \\ compare with \overline{r}=\overline{a}+\lambda \overline{b} \\ \Rightarrow \overline{a}=3\overline{i}-15\overline{j}+9\overline{k} and \overline{b}=2\overline{i}-7\overline{j}+5\overline{k} \\ and \overline{r } =(-\overline{i}+\overline{j}+9\overline{k})+\mu (2\overline{i}+\overline{j}-3\overline{k}) \\ compare with r=\overline{c}+\mu \overline{d} \\ \Rightarrow \overline{c}=-\overline{i}+\overline{j}+9\overline{k} and \overline{d} =2\overline{i}+\overline{j}-3\overline{k} \\ Now \overline{a}-\overline{c}=4\overline{i}-16\overline{\overline{j}} \\ \left[\begin{array}{cc}\overline{a}-\overline{c } \overline{b} \overline{d}& \end{array}\right]=\left|\begin{array}{ccc}4& -16& 0\\ 2& -7& 5\\ 2& 1& -3\end{array}\right| \\ =4(21-5)+16(-6-10)+0 \\ =64-256 \\ =-192 \\ and \overline{b}\times \overline{d} =\left|\begin{array}{ccc}i& j& k\\ 2& -7& 5\\ 2& 1& -3\end{array}\right| \\ =\overline{i}(21-5)-\overline{j}(-6-10)+\overline{k}(2+14) \\ =16 \overline{i }+16 \overline{j}+16 \overline{k} \\ Now \left|\overline{b}\times \overline{d}\right|=\sqrt{{\left(16\right)}^2+{\left(16\right)}^2+{\left(16\right)}^2} \\ =16\sqrt{3} \\ NOw S.D=\left|\frac{\left[\begin{array}{cc}\overline{a}-\overline{c } \overline{b} \overline{d}& \end{array}\right]}{ \left|\overline{b}\times \overline{d}\right|}\right| \\ =\left|\frac{-192}{16\sqrt{3}}\right| \\ =\frac{12}{\sqrt{3}} \\ =4\sqrt{3} \end{gathered}$$