Let $\vec{v} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{w} = \hat{i} + 3\hat{k}$. If $\vec{u}$ is a unit vector,then the maximum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is

The volume of the parallelepiped whose conterminous edges are $\vec{a} = \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} - 4\hat{j} + 5\hat{k}$,and $\vec{c} = 3\hat{i} - 5\hat{j} + 2\hat{k}$ is:

Evaluate: $\vec{a} \cdot \{(\vec{b} + \vec{c}) \times (\vec{a} + \vec{b} + \vec{c})\}$

Let the vectors $\overrightarrow{u}_1 = \hat{i} + \hat{j} + a\hat{k}$,$\overrightarrow{u}_2 = \hat{i} + b\hat{j} + \hat{k}$ and $\overrightarrow{u}_3 = c\hat{i} + \hat{j} + \hat{k}$ be coplanar. If the vectors $\overrightarrow{v}_1 = (a+b)\hat{i} + c\hat{j} + c\hat{k}$,$\overrightarrow{v}_2 = a\hat{i} + (b+c)\hat{j} + a\hat{k}$ and $\overrightarrow{v}_3 = b\hat{i} + b\hat{j} + (c+a)\hat{k}$ are also coplanar,then $6(a+b+c)$ is equal to $..............$.

Let $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{B} = \hat{i}$,and $\overrightarrow{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$. If $C_2 = -1$ and $C_3 = 1$,then to make the three vectors coplanar:

If $\overline{u}, \overline{v}$ and $\overline{w}$ are three non-coplanar vectors,then $(\bar{u}+\bar{v}-\bar{w}) \cdot [(\bar{u}-\bar{v}) \times (\bar{v}-\bar{w})]$ is equal to