Let the position vectors of the vertices \(A, B\) and \(C\) of a tetrahedron \(A B C D\) be \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{k}\) and \(2 \hat{i}+\hat{j}-\hat{k}\) respectively. The altitude from the vertex \(D\) to the opposite face \(A B C\) meets the median line segment through \(A\) of the triangle \(A B C\) at the point \(E\). If the length of \(A D\) 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

1. A \(\frac{1}{12}(7 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+3 \hat{k})\)
2. B \(\frac{1}{2}(\hat{i}+4 \hat{j}+7 \hat{k})\)
3. C \(\frac{1}{6}(12 \hat{i}+12 \hat{j}+\hat{k})\)
4. D \(\frac{1}{6}(7 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}+\hat{\mathrm{k}})\)