Let the position vectors of the vertices $A$, $B$ and $C$ of a tetrahedron $ABCD$ be $\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}$, $\hat{\mathbf{i}} + 3 \hat{\mathbf{j}} - 2 \hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}}$ 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

Step 1: Calculate vectors AB and AC and area of triangle ABC

First, we find the vectors $\vec{AB}$ and $\vec{AC}$ by subtracting the position vector of A from B and C respectively. Then, we calculate the cross product of $\vec{AB}$ and $\vec{AC}$ to find a vector normal to the plane containing triangle ABC. The magnitude of this cross product, divided by 2, gives the area of triangle ABC.

Step 2: Find the height of the tetrahedron

The volume of a tetrahedron is given by the formula $\frac{1}{3} \times \text{base area} \times \text{height}$. Using the given volume and the calculated area of triangle ABC, we can find the height $h$ of the tetrahedron from vertex D to the base ABC. The height is the projection of $\vec{AD}$ onto the normal vector of the plane ABC.

Step 3: Determine the position vector of D

We are given the length of AD. We can set up equations using the dot product of $\vec{AD}$ with the unit normal vector to the plane ABC, which equals the height $h$. We also use the magnitude of $\vec{AD}$. Solving these equations will give us the coordinates of D. For simplicity, we will assume one of the possible solutions for $\vec{d}$.

Step 4: Find the midpoint M of BC and the equation of the median AM

The median line segment through A connects vertex A to the midpoint M of the opposite side BC. We find the position vector of M by averaging the position vectors of B and C. Then, we find the vector $\vec{AM}$ and write the parametric equation of the line AM.

Step 5: Find the point E

The point E is the intersection of the altitude from D and the median AM. The altitude from D is a line passing through D and parallel to the normal vector of the plane ABC. We set up the parametric equations for both lines and equate their components to find the values of the parameters $s$ and $t$. Substituting these values back into either line equation gives the position vector of E.

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