If for vectors $\vec{a}, \vec{b},$ and $\vec{c}$,$\vec{a} + \vec{b} + \vec{c} = \vec{0}$ and $|\vec{a}| = 7, |\vec{b}| = 5, |\vec{c}| = 3$,then the angle between $\vec{b}$ and $\vec{c}$ is ............ $^o$.

Show that the points $A (2 \hat{i}-\hat{j}+\hat{k})$,$B (\hat{i}-3 \hat{j}-5 \hat{k})$,and $C (3 \hat{i}-4 \hat{j}-4 \hat{k})$ are the vertices of a right-angled triangle.

Let $ABCD$ be a parallelogram where $\overrightarrow{AB} = \overrightarrow{a}$,$\overrightarrow{AD} = \overrightarrow{b}$,$|\overrightarrow{a}| = |\overrightarrow{b}| = 2$ and $|\overrightarrow{a} \times \overrightarrow{b}| + \overrightarrow{a} \cdot \overrightarrow{b} = \sqrt{2} |\overrightarrow{a}| |\overrightarrow{b}|$,where $\overrightarrow{a} \cdot \overrightarrow{b} > 0$. Then the area of this parallelogram is (in square units):

If $\overline{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}$,$\overline{b}=2 \hat{i}+\hat{j}-\hat{k}$ and $\overline{c}=\hat{i}+3 \hat{j}$ are such that $(\overline{a}+\lambda \overline{b})$ is perpendicular to $\overline{c}$,then the value of $\lambda$ is

Show that the points $A, B$ and $C$ with position vectors $\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}$,$\vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}$ respectively form the vertices of a right-angled triangle.

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be vectors of equal magnitude such that the angle between $\vec{a}$ and $\vec{b}$ is $\alpha$,$\vec{b}$ and $\vec{c}$ is $\beta$,and $\vec{c}$ and $\vec{a}$ is $\gamma$. Then the minimum value of $\cos \alpha + \cos \beta + \cos \gamma$ is