If $\bar{a} = \bar{i} - 2\bar{j} - 2\bar{k}$ and $\bar{b} = 2\bar{i} + \bar{j} + 2\bar{k}$ are two vectors, then $(\bar{a} + 2\bar{b}) \times (3\bar{a} - \bar{b}) = $

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane defined by $\hat{i}$ and $\hat{i} + \hat{j}$,and the plane defined by $\hat{i} - \hat{j}$ and $\hat{i} + \hat{k}$. Find the angle between $\vec{a}$ and $\hat{i} - 2\hat{j} + 2\hat{k}$.

If $\overrightarrow{A}=i-2j-3k,\,\overrightarrow{B}=2i+j-k,\,\overrightarrow{C}=i+3j-2k$,then $(\overrightarrow A \times \overrightarrow B ) \times \overrightarrow C $ is

Let $\overrightarrow{OA}=\overrightarrow{a}$,$\overrightarrow{OB}=12 \overrightarrow{a}+4 \overrightarrow{b}$,and $\overrightarrow{OC}=\overrightarrow{b}$,where $O$ is the origin. If $S$ is the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$,then the ratio of the area of the quadrilateral $OABC$ to the area of $S$ is:

Let $\overrightarrow{a}=2 \hat{i}-\hat{j}+3 \hat{k}$,$\overrightarrow{b}=3 \hat{i}-5 \hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c}=\vec{c} \times \vec{b}$ and $(\overrightarrow{a}+\overrightarrow{c}) \cdot(\overrightarrow{b}+\overrightarrow{c})=168$. Then the maximum value of $|\vec{c}|^2$ is :

$A$ tetrahedron has vertices $P(1, 2, 1)$,$Q(2, 1, 3)$,$R(-1, 1, 2)$ and $O(0, 0, 0)$. The angle between the faces $OPQ$ and $PQR$ is