Answer:
Given equation of lines can be rewritten as \[\vec{r}=(2\hat{i}-3\hat{j}+5\hat{k})+\lambda (\hat{i}-\hat{j}\,+\hat{k})\] and \[\vec{r}=(-\,\hat{i}-\hat{j}+5\hat{k})+\mu (2\hat{i}\,+4\hat{j}\,-3\hat{k})\] On comparing the above equations with standard vector form of equation of line, \[\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},\] we get \[{{\vec{a}}_{1}}=2\hat{i}-3\hat{j}+5\hat{k},\] \[{{\vec{b}}_{1}}=\hat{i}-\hat{j}+\hat{k},\] \[{{\vec{a}}_{2}}=-\,\hat{i}-\hat{j}+5\hat{k}\] and \[{{\vec{b}}_{2}}=2\hat{i}+4\hat{j}-3\hat{k}\] Now, consider \[\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -\,1 & 1 \\ 2 & 4 & -\,3 \\ \end{matrix} \right|\] \[=\hat{i}(3-4)-\hat{j}(-\,3-2)+\hat{k}(4+2)\] \[\Rightarrow \] \[{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}=-\,\hat{i}+5\hat{j}+6\hat{k}\] \[\Rightarrow \] \[|{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}|\,\,=\sqrt{{{(-1)}^{2}}+{{(5)}^{2}}+{{(6)}^{2}}}\] \[\Rightarrow \] \[=\sqrt{1+25+36}=\sqrt{62}\] Also, \[{{\vec{a}}_{2}}-{{\vec{a}}_{1}}=(-\hat{i}-\hat{j}\,+5\hat{k})-(2\hat{i}-3\hat{j}+5\hat{k})\] \[=-\,3\hat{i}+2\hat{j}\] We know that, shortest distance between two lines is given by \[d=\left| \frac{({{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}).({{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}})}{|{{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}|} \right|\] On purring above values, we get \[d=\left| \frac{(-\hat{i}+5\hat{j}\,+6\hat{k}).(-3\hat{i}\,+2\hat{j})}{\sqrt{62}} \right|\] \[=\left| \frac{3+10+0}{\sqrt{62}} \right|=\frac{13}{\sqrt{62}}=\frac{13\sqrt{62}}{62}\] Hence, required shortest distance is \[\frac{13\sqrt{62}}{62}\] units. Two ships will not met any accidental collision. Value Carelessness leads to mishappening.
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