The distance of the point $3\widehat{i}+5\widehat{k}$ from the line parallel to the vector  $6\widehat{i}+\widehat{j}-2\widehat{k}$ and passing . through the point  $8\widehat{i}+3\widehat{j}+\widehat{k}$ is 

The distance of the point $\overrightarrow{c}$ from the line 

$\overrightarrow{r}=\overrightarrow{a}+s\overrightarrow{b}\text{ is }\frac{\left|\right(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}|}{\left|\overrightarrow{b}\right|}$

$\overrightarrow{c}=3\widehat{i}+5\widehat{k},\overrightarrow{a}=8\widehat{i}+3\widehat{j}+\widehat{k},\overrightarrow{b}=6\widehat{i}+\widehat{j}-2\widehat{k}$

$(\overrightarrow{c}-\overrightarrow{a})\times \overrightarrow{b}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ -5& -3& 4\\ 6& 1& -2\end{array}\right|=2\widehat{i}+14\widehat{j}+13\widehat{k}$

$\text{ Distance }=\frac{\sqrt{2^2+{14}^2+{13}^2}}{\sqrt{6^2+1^2+2^2}}=\sqrt{\frac{369}{41}}=3$