The position vector of the vertices $A,BandC$ of a tetrahedron $ABCD$ are $\widehat{i}+\widehat{j}+\widehat{k},\widehat{i}and3\widehat{i},$ respectively. The altitude from vertex D to the opposite face $ABC$ meets the median line through A of the triangle ABC at a point E. If the length of the side $AD$ is 4 and the volume of the tetrahedron is  $\frac{2\sqrt{2}}{3}$, Then the sum of $x$ co-ordinates of all possible Positions of  E can be

We are given  $AD=4$
Volume of tetrahedron $=\frac{2\sqrt{2}}{3}$
$\therefore \frac{1}{2}|\overrightarrow{BA}\times \overrightarrow{BC}|p=2\sqrt{2}$

$\frac{1}{2}|(\widehat{j}+\widehat{k})\times 2i|p=2\sqrt{2}or|\widehat{j}-\widehat{k}|p=2\sqrt{2}$
Or  $\sqrt{2}p=2\sqrt{2}\therefore p=2$
We have to find the P.V. of point E. Let it divides median AF in the ratio $\lambda :1$
$$\begin{gathered} \therefore P.V.ofEis\frac{\lambda .2\widehat{i}+(\widehat{i}+\widehat{j}+\widehat{k})}{\lambda +1}\mathrm{......}\left(1\right) \\ \therefore \overrightarrow{AE}=P.V.ofE-P.V.ofA=\frac{\lambda }{\lambda +1}(\widehat{i}-\widehat{j}-\widehat{k}) \\ \therefore {\left|\overrightarrow{AE}\right|}^2={\overrightarrow{AE}}^2=\left(\frac{\lambda }{\lambda +1}\right).3\mathrm{......}\left(2\right) \\ Now, p^2+A{E}^2=A{D}^2 \\ Or 4+{\left(\frac{\lambda }{\lambda +1}\right)}^2.3=16\therefore 3\left(\frac{\lambda }{\lambda +1}\right)=12 \\ Or \left(\frac{\lambda }{\lambda +1}\right)=\pm 2 \\ \lambda =\pm (2\lambda +2) \therefore \lambda =-2or-2/3 \end{gathered}$$

Putting the value of  $\lambda$ in  $\left(2\right)$ we get the the P.V. of possible positions of E as  $(-1,3,3)or(3,-1,-1)$