The shortest distance between the lines \(\quad \bar{r}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-2 t) \hat{k}\) and
\(\bar{r}=(p+1) \hat{\imath}+(2 p-1) \hat{\jmath}+(2 p+1) \hat{k} \quad\) is

\(\ell_{1}: \overline{\mathrm{r}} =(1-\mathrm{t}) \hat{\mathrm{i}}+(\mathrm{t}-2) \hat{\mathrm{j}}+(3-2 \mathrm{t}) \hat{\mathrm{k}} \)
\( =(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mathrm{t}(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \)
\( \ell_{2}: \overline{\mathrm{r}} =(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mathrm{p}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \)
\( \text {Here } \overline{\mathrm{a}}_{2}-\overline{\mathrm{a}}_{1}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})-(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
\(\overline{\text b }_1 \times \overline{\text b }_2=\left|\begin{array}{ccc}\hat{\text i } & \hat{\text j } & \hat{\text k } \\ -1 & 1 & -2 \\ 1 & 2 & 2\end{array}\right|=\hat{\text i }(2+4)-\hat{\text j }(0)-\) \(3 \hat{\text k }=6 \hat{\text i }-3 \hat{\text k }\)
\(\therefore\left|\bar{b}_{1} \times \bar{b}_{2}\right| \quad=\sqrt{36+9}=3 \sqrt{5}\)
shortest distance \(=\left|\frac{\left(\bar{b}_{1} \times \bar{b}_{2}\right) \cdot\left(\bar{a}_{2}-\bar{a}_{1}\right)}{\left(\bar{b}_{1} \times \bar{b}_{2}\right)}\right|=\left|\frac{(6 \hat{i}-3 \hat{k}) \cdot(\hat{j}-2 \hat{k})}{3 \sqrt{5}}\right|=\frac{6}{3 \sqrt{5}}=\frac{2}{\sqrt{5}}\)