Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$. If $\vec{b}$ is a vector such that $\vec{a} \cdot \vec{b} = |\vec{b}|^2$ and $|\vec{a} - \vec{b}| = \sqrt{7}$,then find $|\vec{b}|$.

$\overline{u}, \overline{v}, \overline{w}$ are three vectors such that $|\overline{u}|=1, |\overline{v}|=2, |\overline{w}|=3$. If the projection of $\overline{v}$ along $\overline{u}$ is equal to the projection of $\overline{w}$ along $\overline{u}$ and $\overline{v}, \overline{w}$ are perpendicular to each other,then $|\overline{u}-\overline{v}+\overline{w}|=$

If the position vectors of $A, B, C, D$ are $\bar{i}+2\bar{j}+2\bar{k}, 2\bar{i}-\bar{j}, \bar{i}+\bar{j}+3\bar{k}$ and $4\bar{j}+5\bar{k}$ respectively,then the quadrilateral $ABCD$ is a

If $P=(0,1,2)$, $Q=(4,-2,1)$, and $O=(0,0,0)$, then $\angle POQ$ is equal to

$(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$ if and only if . . . . . . (where $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$).

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: