If in a right-angled triangle $ABC$,the hypotenuse $|\overrightarrow{AB}| = p$,then $\overrightarrow{AB} \cdot \overrightarrow{AC} + \overrightarrow{BC} \cdot \overrightarrow{BA} + \overrightarrow{CA} \cdot \overrightarrow{CB} = $

Let the vectors $\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$. For some $\lambda, \mu \in \mathbb{R}$, let $\vec{c} = \lambda \vec{a} + \mu \vec{b}$. If $\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10$ and $\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2$, then $|\vec{c}|^2$ is equal to:

$A$ unit vector in the $xy$-plane that makes an angle of $45^{\circ}$ with the vector $(i + j)$ and an angle of $60^{\circ}$ with the vector $(3i - 4j)$ is:

Two adjacent sides of a parallelogram $ABCD$ are given by $\overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $AD$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $AD$ becomes $AD'$. If $AD'$ makes a right angle with the side $AB$,then the cosine of the angle $\alpha$ is given by

The shortest distance between the lines $r = 3i + 5j + 7k + \lambda(i + 2j + k)$ and $r = -i - j - k + \mu(7i - 6j + k)$ is

Three vectors $\vec a, \vec b, \vec c$ are inclined at an acute angle with each other such that $|\vec a| = 2, |\vec b| = 3, |\vec c| = 9$ and the lengths of the projections of $\vec a$ on $\vec b$,$\vec b$ on $\vec c$,and $\vec c$ on $\vec a$ respectively are in geometric progression. If the angle between $\vec a$ and $\vec b$ is $\frac{5\pi}{12}$ and the angle between $\vec c$ and $\vec a$ is $\frac{\pi}{12}$,then the angle between $\vec b$ and $\vec c$ is: