The vector of magnitude $3\sqrt{6}$ along the bisector of the angle between the vectors $4\overline{\mathrm{i}}-7\overline{\mathrm{j}}+4\overline{\mathrm{k}} \mathrm{and} \overline{\mathrm{i}}+2\overline{\mathrm{j}}-2\overline{\mathrm{k}}$

$\overline{a}=4\overline{i}-7\overrightarrow{j}+4\overrightarrow{k}$

$\left|\overrightarrow{a}\right|=\sqrt{16+49+16}=9$

$\overline{b}=\overline{i}+2\overrightarrow{j}-2\overrightarrow{k}$

$\left|b\right|=\sqrt{1+4+4}=3$

$$\begin{gathered} \hat{\overline{a}}=\frac{\overline{a}}{\left|\overline{a}\right|}=\frac{4\overrightarrow{i}-7\overrightarrow{j}+4\overrightarrow{k}}{9} \\ \hat{\overline{b}}=\frac{\overline{b}}{\left|\overline{b}\right|}=\frac{\overrightarrow{i}+2\overline{j}-2\overline{k}}{3} \\ \overline{b}=\overline{i}+2\overrightarrow{j}-2\overrightarrow{k} \\ \left|\overrightarrow{b}\right|=\sqrt{1+4+4}=3 \\ \hat{\overline{a}}=\frac{\overline{a}}{\left|\overrightarrow{a}\right|}=\frac{4\overrightarrow{i}-7\overrightarrow{j}+4\overrightarrow{k}}{9} \\ \hat{\overline{b}}=\frac{\hat{b}}{\left|\overline{b}\right|}=\frac{\overrightarrow{i}+2\overline{j}-2\overline{k}}{3} \\ \pm 3\sqrt{6}\left(\frac{\hat{\overline{a}}+\hat{\overline{b}}}{|\hat{\overline{a}}+\hat{\overline{b}}|}\right) \\ \hat{\overrightarrow{a}}+\hat{\overline{b}}=\frac{4\overrightarrow{i}-7\overrightarrow{j}+4\overrightarrow{k}}{9}+\frac{\overrightarrow{i}+\overrightarrow{y}-2\overline{k}}{3} \\ =\pm 3\sqrt{6}\left(\frac{4\overrightarrow{i}-7\overrightarrow{j}+4\overrightarrow{k}}{9}+\frac{\overrightarrow{i}+2\overrightarrow{j}-2\overrightarrow{k}}{3}\right) \\ =\frac{4\overrightarrow{i}-7\overrightarrow{j}+4\overrightarrow{k}+3\overrightarrow{i}+6\overrightarrow{j}-6\overrightarrow{k}}{9} \\ =\frac{7\overrightarrow{i}-\overrightarrow{j}-2\overrightarrow{k}}{9} \\ \therefore \pm (7\overrightarrow{i}-\overrightarrow{j}-2\overrightarrow{k}) \end{gathered}$$