$\vec{a}, \vec{b}, \text{ and } \vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$. If $\vec{a}, \vec{b}, \vec{c}$ are perpendicular to the vectors $\vec{b}+\vec{c}, \vec{c}+\vec{a}, \vec{a}+\vec{b}$ respectively,then $\sqrt{|\vec{a}+\vec{b}+\vec{c}|^2-2} = $

If the angle between $\vec{a}$ and $\vec{b}$ is $\frac{2\pi}{3}$ and the projection of $\vec{a}$ in the direction of $\vec{b}$ is $-2$,then find $|\vec{a}|$.

The value of $\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})$ is

If $A(3,4,5), B(4,6,3), C(-1,2,4)$ and $D(1,0,5)$ are such that the angle between the lines $DC$ and $AB$ is $\theta$,then $\cos \theta$ is equal to

If $\bar{a}, \bar{b}$ and $\bar{c}$ are vectors such that $|\bar{a}| = |\frac{\bar{b}}{2}| = |\frac{\bar{c}}{3}| = 1$; $\bar{b}$ and $\bar{c}$ are perpendicular; and the projections of $\bar{b}$ and $\bar{c}$ on $\bar{a}$ are equal,then $|\bar{a} - \bar{b} + \bar{c}| = $

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