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 through $A$ of $\triangle ABC$ at the point $E$. If the length of $AD$ is $\dfrac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\dfrac{\sqrt{805}}{6\sqrt{2}}$, then the position vector of $E$ is: