Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} - \hat{j} - \hat{k}$ be three vectors. $A$ vector $\vec{v}$ in the plane of $\vec{a}$ and $\vec{b}$,whose projection on $\vec{c}$ is $1/\sqrt{3}$,is given by:

The projection of $\overrightarrow{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$ on $\overrightarrow{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ is

If $|\vec{a}| = \sqrt{27}$,$|\vec{b}| = 7$ and $|\vec{a} \times \vec{b}| = 35$,then $\vec{a} \cdot \vec{b}$ is equal to

The angle between the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ is

$ABC$ is a right-angled triangle in which $\max \{AB, BC, AC\} = BC$. If the position vectors of $B$ and $C$ are respectively $3\hat{i}-2\hat{j}+\hat{k}$ and $5\hat{i}+\hat{j}-3\hat{k}$,then find the value of $AB \cdot AC + BA \cdot BC + CA \cdot CB$.

Let a unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$ make angles $\frac{\pi}{2}, \frac{\pi}{3}$ and $\frac{2 \pi}{3}$ with the vectors $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$ and $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ respectively. If $\overrightarrow{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$,then $|\hat{u}-\overrightarrow{v}|^2$ is equal to