If $u = 2i + 2j - k$ and $v = 6i - 3j + 2k$,then find the unit vector perpendicular to both $u$ and $v$.

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors such that $\vec{a} \times \vec{b} = 4\vec{c}$,$\vec{b} \times \vec{c} = 9\vec{a}$,and $\vec{c} \times \vec{a} = \alpha\vec{b}$,where $\alpha > 0$. If $|\vec{a}| + |\vec{b}| + |\vec{c}| = 36$,then $\alpha$ is equal to:

The area of a triangle with vertices $(1, 2, 0)$,$(1, 0, a)$,and $(0, 3, 1)$ is $\sqrt{6}$ sq. units. Then the values of '$a$' are:

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k},$ find a vector $\vec{c}$ such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3.$

The area of the parallelogram whose diagonals are represented by the vectors $\bar{a}=3 \hat{i}-\hat{j}-2 \hat{k}$ and $\bar{b}=-\hat{i}+3 \hat{j}-3 \hat{k}$ is

Let $\overrightarrow{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\hat{i}-2 \hat{j}-3 \hat{k}$ be two vectors. If $A_1$ is the area of the quadrilateral having $\vec{a}, \vec{b}$ as its diagonals and $A_2$ is the area of the parallelogram having $\overrightarrow{a}, \overrightarrow{b}$ as its two adjacent sides,then $A_1 \cdot A_2=$