The distance of the point having position vector \(\hat{i}-2 \hat{j}-6 \hat{k}\), from the straight line passing through the point \((2,-3,-4)\) and parallel to the vector \(6 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}\) is units.

Given equation of the line is
\(\begin{aligned}
& \overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}} \\
& \therefore \quad \bar{r}=2 \hat{i}-3 \hat{j}-4 \hat{k}+\lambda(6 \hat{i}+3 \hat{j}-4 \hat{k}) \\
&
\end{aligned}\)
To find: Its distance from point \(\bar{\alpha}=\hat{i}-2 \hat{j}-6 \hat{k}\).
\(\therefore \quad \text { Required distance }=\sqrt{|\bar{\alpha}-\bar{a}|^2-\left[\frac{(\bar{\alpha}-\bar{a}) \cdot \bar{b}}{|\bar{b}|}\right]^2}\)
Here, \(|\bar{\alpha}-\bar{a}|^2=6\) and \(\left[\frac{(\bar{\alpha}-\bar{a}) \cdot \bar{b}}{|\bar{b}|}\right]^2=\frac{25}{61}\).
\(\therefore \quad \text { Requiređ distance }=\sqrt{\frac{341}{61}}\)