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 :

The given lines are <br/><br/>$\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}$ and $\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}$ <br/><br/>$$ \begin{aligned} & \text { So, } \vec{b}_1=4 \hat{i}+5 \hat{j}+3 \hat{k} \\\\ & \vec{b}_2=3 \hat{i}+4 \hat{j}+2 \hat{k} \\\\ & \vec{a}_1=4 \hat{i}-2 \hat{j}-3 \hat{k} \\\\ &\vec{a}_2=\hat{i}+3 \hat{j}+4 \hat{k} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \therefore \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{array}\right| \\\\ & =(10-12) \hat{i}-(8-9) \hat{j}+(16-15) \hat{k} \\\\ & =-2 \hat{i}+\hat{j}+\hat{k} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { Shortest distance, } d=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right| \\\\ & =\left|\frac{(3 \hat{i}-5 \hat{j}-7 \hat{k}) \cdot(-2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{4+1+1}}\right| \\\\ & =\left|\frac{-6-5-7}{\sqrt{6}}\right|=\frac{18}{\sqrt{6}}=3 \sqrt{6} \text { units } \end{aligned} $$