Distance between the parallel lines \(\frac{x}{3}=\frac{y-1}{-2}=\frac{z}{1}\) and \(\frac{x+4}{3}=\frac{y-3}{-2}=\frac{z+2}{1}\) is

The vector equations of the given lines are
\(\begin{aligned}
& \overline{\mathrm{r}}=\hat{\mathrm{j}}+\lambda(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \text { and } \\
& \overline{\mathrm{r}}=-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}+\mu(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})
\end{aligned}\)
The distance between the parallel lines \(\bar{r}=\bar{a}_1+\lambda \bar{b}\) and \(\bar{r}=\bar{a}_2+\mu \bar{b}\) is given by
\(d=\left|\frac{\left(\bar{a}_2-\bar{a}_1\right) \times \bar{b}}{|\bar{b}|}\right|\)
\(\text {Here, } \overline{\mathrm{a}}_1=\hat{\mathrm{j}}, \overline{\mathrm{a}}_2=-4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \)
\( \therefore \overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1=-4 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
\(\begin{aligned}
& \left(\bar{a}_2-\bar{a}_1\right) \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
-4 & 2 & -2 \\
3 & -2 & 1
\end{array}\right| \\
& =-2 \hat{i}-2 \hat{j}+2 \hat{k} \\
& |\bar{b}|=\sqrt{9+4+1}=\sqrt{14}
\end{aligned}\)
\(\begin{aligned} \therefore \quad d & =\left|\frac{-2 \hat{i}-2 \hat{j}+2 \hat{k}}{\sqrt{14}}\right| \\ & =\sqrt{\frac{4+4+4}{14}} \\ & =\sqrt{\frac{12}{14}}=\sqrt{\frac{6}{7}} \text { units }\end{aligned}\)