Area of $\triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$

$$=\frac{1}{2}|5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}|=\frac{1}{2} \sqrt{35}$$

volume of tetrahedron

$$\begin{aligned} & =\frac{1}{3} \times \text { Base area } \times \mathrm{h}=\frac{\sqrt{805}}{6 \sqrt{2}} \\ & \frac{1}{3} \times \frac{1}{2} \sqrt{35} \times \mathrm{h}=\frac{\sqrt{805}}{6 \sqrt{2}} \\ & \mathrm{~h}=\sqrt{\frac{23}{2}} \end{aligned}$$

$$\mathrm{AE}^2=\mathrm{AD}^2-\mathrm{DE}^2=\frac{13}{18} \therefore \mathrm{AE}=\sqrt{\frac{13}{18}}$$

$$\begin{aligned} & \overrightarrow{\mathrm{AE}}=|\mathrm{AE}| \cdot\left(\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{\sqrt{26}}\right) \\ & =\sqrt{\frac{13}{18}} \cdot\left(\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{\sqrt{26}}\right) \\ & =\sqrt{\frac{13}{18}} \cdot\left(\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{\sqrt{26}}\right)=\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{6} \\ & \text { P.V. of } \mathrm{E}=\frac{\hat{\mathrm{i}}-5 \hat{\mathrm{k}}}{6}+\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}=\frac{1}{6}(7 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \end{aligned}$$