If $a$,$b$,and $c$ are unit vectors,then $|a - b|^2 + |b - c|^2 + |c - a|^2$ does not exceed

Consider two vectors $\overrightarrow{u} = 3\hat{i} - \hat{j}$ and $\overrightarrow{v} = 2\hat{i} + \hat{j} - \lambda\hat{k}$,where $\lambda > 0$. The angle between them is given by $\cos^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right)$. Let $\vec{v} = \vec{v}_1 + \vec{v}_2$,where $\vec{v}_1$ is parallel to $\overrightarrow{u}$ and $\vec{v}_2$ is perpendicular to $\overrightarrow{u}$. Then the value $|\vec{v}_1|^2 + |\vec{v}_2|^2$ is equal to

If $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$,$\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ and the length of the internal bisector of $\angle BOA$ of triangle $AOB$ is $k$,then $9k^2 =$

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{14}$ and $\vec{a} \cdot \vec{b}=-7$,then $\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}=$

For any two vectors $\vec{a}$ and $\vec{b}$,$|\vec{a}| |\vec{b}|$ . . . . . . $|\vec{a} \cdot \vec{b}|$.

If $\overrightarrow{A} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\overrightarrow{B} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{C} = 3\hat{i} + \hat{j}$,then the value of $t$ such that $\overrightarrow{A} + t\overrightarrow{B}$ is at a right angle to vector $3\hat{i} + 4\hat{j}$ is