Area (in sq. units) of the triangle with vertices A (1,1,1) B (1,2,3) and C (2,3,1) is:

$[\vec{a} + \vec{b},\ \vec{b} + \vec{c}, \vec{a} + \vec{b} + \vec{c}]$ is equal to

Vectors $\vec{a}, \vec{b}, \vec{c}$ are such that $\vec{a} + \vec{b} + \vec{c} = 0$ and $|\vec{a}| = |\vec{b}| = |\vec{c}|$ then, the angle between any two of them, is

The vectors from origin to the points A and B are $\vec{a}=2\hat{i}-3\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+3\hat{j}+\hat{k}$ respectively, then the area (in Sq. unit) of triangle OAB is:

The position vector of the point which divides the join of the points $2\vec{a}-3\vec{b}$ and $3\vec{a}-2\vec{b}$ externally in the ratio 2:3, is:

If $\vec{a} = 2\hat{j} - \hat{k}$, $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$ and $\vec{c} = -\hat{i} + \hat{k}$ are three vectors, then the area (in sq. units) of the parallelogram whose diagonals are $(\vec{b} + \vec{c})$ and $(\vec{a} + \vec{c})$ is

If $\vec{a}$ and $\vec{b}$ are two non-zero vectors such that $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$, then the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is