If $\hat{a}, \hat{b}, \hat{c}$ are unit vectors,then the least value of $|\hat{a}+\hat{b}|^2+|\hat{b}+\hat{c}|^2+|\hat{c}+\hat{a}|^2$ will be-

Let $a = 2i - j + k$,$b = i + 2j - k$,and $c = i + j - 2k$ be three vectors. $A$ vector in the plane of $b$ and $c$ whose projection on $a$ is of magnitude $\sqrt{2/3}$ is

Let $\vec{a}, \vec{b}, \vec{c}$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $\theta$ with the vector $\vec{a}+\vec{b}+\vec{c}$. Then $36 \cos ^{2} 2 \theta$ is equal to $.....$

$3 \hat{i}-2 \hat{j}-\hat{k}, -2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle ABC$ respectively. If $H$ is its orthocenter,then $\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC} = $

The value of $\frac{(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2}{2|\vec{a}|^2|\vec{b}|^2}$ is

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