We have $\overrightarrow{a}=\mathrm{ }2\hat{\mathrm{i}} + \widehat{\mathrm{j}\mathrm{ }}-\mathrm{ }\hat{\mathrm{k}}$

             $\overrightarrow{b}=\mathrm{ }\hat{\mathrm{i}} -2 \widehat{\mathrm{j}\mathrm{ }}+\mathrm{ }\hat{\mathrm{k}}$

Internal and External angle bisecting Vectors  between $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by  $\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|} + \frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$ and  $\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|} - \frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}$ respectively.

Internal angle bisector= $\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|} + \frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\overrightarrow{p}\left(say\right)$

$\Rightarrow \overrightarrow{p}=\left(\frac{\overrightarrow{a} + \overrightarrow{b}}{\sqrt{6}}\right)$

$\Rightarrow \overrightarrow{p}=\left(\frac{3\hat{\mathrm{i}} - \widehat{\mathrm{j}\mathrm{ }}}{\sqrt{6}}\right)$

External Angle bisector=

$=\left(\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|} - \frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}\right)=\overrightarrow{q}\left(say\right)$

$\Rightarrow \overrightarrow{q}=\left(\frac{\overrightarrow{a} - \overrightarrow{b}}{\sqrt{6}}\right)$

$$$\Rightarrow \overrightarrow{q}=\left(\frac{\hat{\mathrm{i}} +3 \widehat{\mathrm{j}\mathrm{ }}-2 \hat{\mathrm{k}}}{\sqrt{6}}\right)$

Since $\overrightarrow{p}.\overrightarrow{b}>0$ and $\overrightarrow{q}.\overrightarrow{b}<0$

hence required vector will along $\overrightarrow{q}$.

Let required vector be

$\overrightarrow{r}=k\overrightarrow{q}$

Since  $\left|\overrightarrow{r}\right|=3$

$\Rightarrow k=\frac{3\sqrt{6}}{\sqrt{14}}$

Hence required vector

$\Rightarrow \overrightarrow{r}=3\left(\frac{\hat{\mathrm{i}} +3 \widehat{\mathrm{j}\mathrm{ }}-2 \hat{\mathrm{k}}}{\sqrt{14}}\right)$