The shortest distance between the lines $\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$ and $\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$ is  

Given:

$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$

$\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$

So,

$$\begin{gathered} {\overrightarrow{a}}_1=4\hat{i}-2\hat{j}-3\hat{k} \\ {\overrightarrow{a}}_2=\hat{i}+3\hat{j}+4\hat{k} \end{gathered}$$

So,

${\overrightarrow{a}}_2-{\overrightarrow{a}}_1=-3\hat{i}+5\hat{j}+7\hat{k}$

And,

${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 5& 3\\ 3& 4& 2\end{array}\right|=-2\hat{i}+\hat{j}+\hat{k}$

$\Rightarrow \left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|=\sqrt{{\left(-2\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}=\sqrt{6}$

Shortest distance between the lines 

$=\left|\frac{\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)·\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right)}{\left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|}\right|$

$=\left|\frac{\left(-3\hat{i}+5\hat{j}+7\hat{k}\right)·\left(-2\hat{i}+\hat{j}+\hat{k}\right)}{\sqrt{6}}\right|$

$=\left|\frac{6+5+7}{\sqrt{6}}\right|=\frac{6\times 3}{\sqrt{6}} \mathrm{units}$

$=3\sqrt{6} \mathrm{units}$

Hence this is the correct option.