If the position vectors of the vertices $A, B$ and $C$ are $6i$,$6j$ and $k$ respectively with respect to the origin $O$,then the volume of the tetrahedron $OABC$ is

If the vectors $a \hat{i}+\hat{j}+\hat{k}$,$\hat{i}+b \hat{j}+\hat{k}$,and $\hat{i}+\hat{j}+c \hat{k}$ are coplanar,where $(a, b, c \neq 1)$,then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$

The four points whose position vectors are given by $2\bar{a}+3\bar{b}-\bar{c}$, $\bar{a}-2\bar{b}+3\bar{c}$, $3\bar{a}+4\bar{b}-2\bar{c}$ and $\bar{a}-6\bar{b}+6\bar{c}$ are

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\vec{c} = x\hat{i} + (x-2)\hat{j} - \hat{k}$. If the vector $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$,then $x$ equals:

If $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar vectors and $p, q$ are real numbers,then the equality $[3\vec{u}, p\vec{v}, p\vec{w}] - [p\vec{v}, \vec{w}, q\vec{u}] - [2\vec{w}, q\vec{v}, q\vec{u}] = 0$ holds for which of the following?

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors,then $\left| {\begin{array}{*{20}{c}} {\vec{a} \cdot \vec{a}} & {\vec{a} \cdot \vec{b}} & {\vec{a} \cdot \vec{c}} \\ {\vec{b} \cdot \vec{a}} & {\vec{b} \cdot \vec{b}} & {\vec{b} \cdot \vec{c}} \\ {\vec{c} \cdot \vec{a}} & {\vec{c} \cdot \vec{b}} & {\vec{c} \cdot \vec{c}} \end{array}} \right| = \dots$