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

$\bar{a}, \bar{b}, \bar{c}$ are unit vectors. If $\bar{a}, \bar{b}$ are perpendicular vectors,$(\bar{a}-\bar{c}) \cdot(\bar{b}+\bar{c})=0$ and $\bar{c}=l \bar{a}+m \bar{b}+n(\bar{a} \times \bar{b})$ ($l, m, n$ are scalars),then $n^2=$

Let $a = 2\hat{i} - \hat{j} + 2\hat{k}$ and $b = 3\hat{i} - 2\hat{j} - 5\hat{k}$ be two vectors. Then the projection vector of $b$ on a vector perpendicular to $a$ is

$A$ tetrahedron has vertices at $O(0, 0, 0)$,$A(1, 2, 1)$,$B(2, 1, 3)$,and $C(-1, 1, 2)$. The angle between the faces $OAB$ and $ABC$ is:

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors,then the greatest value of $\sqrt{3}|\overrightarrow{a}+\overrightarrow{b}|+|\overrightarrow{a}-\overrightarrow{b}|$ is

If the vectors $\vec{AB} = p \hat{i} + q \hat{j} + r \hat{k}$, $\vec{AC} = s \hat{i} + 3 \hat{j} + 4 \hat{k}$, and $\vec{CB} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ form a $\triangle ABC$, then the values of $p, q, r$ and $s$ such that the area of that $\triangle ABC$ is $5 \sqrt{6}$ are: