Given, lines are $\frac{x-7}{3}=\frac{y+4}{-16}=\frac{z-6}{7}$ and $\frac{x-10}{3}=\frac{y-30}{8}=\frac{4-z}{5}$
The vector form of given lines are $\mathrm{r}=7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}+\lambda(3 \hat{\mathrm{i}}-16 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})$ and $\mathrm{r}=10 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}+\mu(3 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$
On comparing these equations with $\mathrm{r}=\mathrm{a}_{1}+\lambda \mathrm{b}_{1}$ and $\mathrm{r}=\mathrm{a}_{2}+\mu \mathrm{b}_{2}$, we get
$\begin{array}{l}
\overrightarrow{\mathrm{a}}_{1}=7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} \\
\overrightarrow{\mathrm{a}}_{2}=10 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}
\end{array}$
$\overrightarrow{\mathrm{b}}_{1}=3 \hat{\mathrm{i}}-16 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$
and $\overrightarrow{\mathrm{b}}_{2}=3 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}$
Shortest distance $=\left|\frac{\left(\overrightarrow{\mathrm{a}}_{2}-\overrightarrow{\mathrm{a}}_{1}\right) \cdot\left(\overrightarrow{\mathrm{b}}_{1} \times \overrightarrow{\mathrm{b}}_{2}\right)}{\left|\mathrm{b}_{1} \times \mathrm{b}_{2}\right|}\right|$
$=\left|\frac{(3 \hat{\mathrm{i}}+34 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot(24 \hat{\mathrm{i}}+36 \hat{\mathrm{j}}+72 \hat{\mathrm{k}})}{84}\right|$
$=\left|\frac{72+1224-144}{84}\right|=\left|\frac{1152}{84}\right|=\frac{288}{21}$ units