Let the position vectors of the vertices A, B and C of a tetrahedron ABCD be $$\widehat{i}+2\widehat{j}+\widehat{k},\widehat{i}+3\widehat{j}-2\widehat{k}$$ and $$2\widehat{i}+\widehat{j}-\widehat{k}$$ respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through of the triangle ABC at the point . 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 is

The position vectors of vertices A, B, and C are given as: $$\overrightarrow{A} = \widehat{i} + 2\widehat{j} + \widehat{k},\quad \overrightarrow{B} = \widehat{i} + 3\widehat{j} - 2\widehat{k},\quad \overrightarrow{C} = 2\widehat{i} + \widehat{j} - \widehat{k}.$$

The median from vertex A to side BC is found by determining the midpoint M of BC: $$\overrightarrow{M} = \frac{\overrightarrow{B} + \overrightarrow{C}}{2} = \frac{(\widehat{i} + 3\widehat{j} - 2\widehat{k}) + (2\widehat{i} + \widehat{j} - \widehat{k})}{2} = \frac{3\widehat{i} + 4\widehat{j} - 3\widehat{k}}{2} = \frac{3}{2}\widehat{i} + 2\widehat{j} - \frac{3}{2}\widehat{k}.$$ The vector $$\overrightarrow{AM}$$ is $$\overrightarrow{AM} = \overrightarrow{M} - \overrightarrow{A} = \Bigl(\tfrac{3}{2}\widehat{i} + 2\widehat{j} - \tfrac{3}{2}\widehat{k}\Bigr) - (\widehat{i} + 2\widehat{j} + \widehat{k}) = \tfrac{1}{2}\widehat{i} - \tfrac{5}{2}\widehat{k}.$$

The parametric equation of the median AM is $$\overrightarrow{P}(t) = \overrightarrow{A} + t\,\overrightarrow{AM} = (\widehat{i} + 2\widehat{j} + \widehat{k}) + t\Bigl(\tfrac{1}{2}\widehat{i} - \tfrac{5}{2}\widehat{k}\Bigr) = \Bigl(1 + \tfrac{t}{2}\Bigr)\widehat{i} + 2\widehat{j} + \Bigl(1 - \tfrac{5t}{2}\Bigr)\widehat{k}.$$

The altitude from D to the plane ABC is perpendicular to the plane. The normal vector to plane ABC is $$\overrightarrow{AB}\times\overrightarrow{AC}$$, where $$\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A} = (\widehat{i} + 3\widehat{j} - 2\widehat{k}) - (\widehat{i} + 2\widehat{j} + \widehat{k}) = \widehat{j} - 3\widehat{k},$$ $$\overrightarrow{AC} = \overrightarrow{C} - \overrightarrow{A} = (2\widehat{i} + \widehat{j} - \widehat{k}) - (\widehat{i} + 2\widehat{j} + \widehat{k}) = \widehat{i} - \widehat{j} - 2\widehat{k}.$$ Their cross product is $$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 0 & 1 & -3 \\ 1 & -1 & -2 \end{vmatrix} = \widehat{i}(-2 - 3) - \widehat{j}(0 + 3) + \widehat{k}(0 - 1) = -5\widehat{i} - 3\widehat{j} - \widehat{k}.$$ Thus the direction of the altitude is parallel to $$\overrightarrow{N} = -5\widehat{i} - 3\widehat{j} - \widehat{k}$$, and its parametric equation from D is $$\overrightarrow{Q}(s) = \overrightarrow{D} + s\,\overrightarrow{N} = (x\widehat{i} + y\widehat{j} + z\widehat{k}) + s(-5\widehat{i} - 3\widehat{j} - \widehat{k}) = (x - 5s)\widehat{i} + (y - 3s)\widehat{j} + (z - s)\widehat{k}.$$

The intersection point of the altitude and the median satisfies $$(x - 5s)\widehat{i} + (y - 3s)\widehat{j} + (z - s)\widehat{k} = \Bigl(1 + \tfrac{t}{2}\Bigr)\widehat{i} + 2\widehat{j} + \Bigl(1 - \tfrac{5t}{2}\Bigr)\widehat{k},$$ so equating components gives $$x - 5s = 1 + \tfrac{t}{2},\quad y - 3s = 2,\quad z - s = 1 - \tfrac{5t}{2}.$$ From $$y - 3s = 2$$ we get $$y = 2 + 3s$$. Defining $$u = x - 1,\ v = y - 2,\ w = z - 1$$, then $$v = 3s$$. From $$x - 5s = 1 + t/2$$ we have $$u - 5s = t/2\implies t = 2(u - 5s)$$, and from $$z - s = 1 - 5t/2$$ we obtain $$w - s = -5t/2$$. Substituting $$t = 2(u - 5s)$$ yields $$w - s = -5(u - 5s)\implies w = -5u + 26s.$$

The volume of the tetrahedron is $$\tfrac{\sqrt{805}}{6\sqrt{2}}$$, so the scalar triple product satisfies $$\bigl|\overrightarrow{AD}\cdot(\overrightarrow{AB}\times\overrightarrow{AC})\bigr| = \bigl|-5u - 3v - w\bigr| = \frac{\sqrt{805}}{\sqrt{2}} = \sqrt{\tfrac{805}{2}}.$$ Substituting $$v = 3s$$ and $$w = -5u + 26s$$ gives $$S = -5u - 9s + 5u - 26s = -35s,\quad |S| = 35|s| = \sqrt{\tfrac{805}{2}},$$ so $$1225s^2 = \frac{805}{2}\implies s^2 = \frac{23}{70}.$$

The distance $$AD$$ is $$\tfrac{\sqrt{110}}{3}$$, hence $$|\overrightarrow{AD}|^2 = u^2 + v^2 + w^2 = u^2 + 9s^2 + (-5u + 26s)^2 = \frac{110}{9}.$$ Expanding yields $$26u^2 - 260us + 685s^2 = \frac{110}{9},$$ and substituting $$s^2 = \frac{23}{70}$$ gives the quadratic $$252u^2 - 2520su + 2063 = 0.$$

Solving for $$u$$: $$u = \frac{2520s \pm \sqrt{(2520s)^2 - 4\cdot252\cdot2063}}{504} = \frac{2520s \pm 84}{504} = 5s \pm \tfrac{1}{6}.$$ Thus $$u = 5s + \tfrac{1}{6}$$ or $$u = 5s - \tfrac{1}{6}$$, giving $$t = \tfrac{1}{3}$$ or $$t = -\tfrac{1}{3}$$ respectively.

The intersection point $$P$$ on the median is then: for $$t = \tfrac{1}{3}$$, $$\overrightarrow{P} = \tfrac{1}{6}(7\widehat{i} + 12\widehat{j} + \widehat{k});$$ for $$t = -\tfrac{1}{3}$$, $$\overrightarrow{P} = \tfrac{1}{6}(5\widehat{i} + 12\widehat{j} + 11\widehat{k}).$$ Comparing with the options shows that Option D, $$\tfrac{1}{6}(7\widehat{i} + 12\widehat{j} + \widehat{k})$$, matches the case $$t = \tfrac{1}{3}$$. Verification with $$AD$$ length and volume confirms consistency.

Final Answer: D. $$\frac{1}{6}\left(7\widehat{i}+12\widehat{j}+\widehat{k}\right)$$