A vector of magnitude 3, bisecting the angle between the vectors  $\overrightarrow{a}=2\widehat{i}+\widehat{j}-\widehat{k}$  and  $\overrightarrow{b}=\widehat{i}-2\widehat{j}+\widehat{k}$and making an obtuse angle with $\overrightarrow{b}$  is  $\frac{\mathcal{l}(\widehat{i}+\mathcal{l}\widehat{j}-m\widehat{k})}{\sqrt{n}}$, where  $\mathcal{l}+m+n=\_\_\_$

A vector bisecting the angle between $\overrightarrow{a}$  and  $\overrightarrow{b}$  is  $\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}\pm \frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$ ;
in the case  $\frac{2\widehat{i}+\widehat{j}-\widehat{k}}{\sqrt{6}}\pm \frac{\widehat{i}-2\widehat{j}+\widehat{k}}{\sqrt{6}}$
i.e., $\frac{3\widehat{i}-\widehat{j}}{\sqrt{6}}or\frac{\widehat{i}+3\widehat{j}-2\widehat{k}}{\sqrt{6}}$
A vector of magnitude 3 along these vector is  $\frac{3(3\widehat{i}-\widehat{j})}{\sqrt{10}}or\frac{3(\widehat{i}+3\widehat{j}-2\widehat{k})}{\sqrt{14}}$
Now, $\frac{3}{\sqrt{14}}(\widehat{i}+3\widehat{j}-2\widehat{k}).(\widehat{i}-2\widehat{j}+\widehat{k})$ is negative and hence $\frac{3}{\sqrt{14}}(\widehat{i}+3\widehat{j}-2\widehat{k})$  makes an obtuse angle with  $\overrightarrow{b}$.