Let the position vectors of the vertices $A, B$ and $C$ of a tetrahedron $ABCD$ be $\hat{i} + 2\hat{j} + \hat{k}$, $\hat{i} + 3\hat{j} - 2\hat{k}$ and $2\hat{i} + \hat{j} - \hat{k}$ respectively. The altitude from the vertex $D$ to the opposite face $ABC$ meets the median line segment through $A$ of the triangle $ABC$ at the point $E$. If the length of $AD$ is $\frac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6\sqrt{2}}$, then the position vector of $E$ is:
