If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$ represent the adjacent sides of a parallelogram,then the area of this parallelogram is:

If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,find $\vec{a} \times \vec{b}$.

If the vertices of a triangle are $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$,then the area of the triangle is:

Let $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}$,$\vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}$ and a vector $\vec{c}$ be such that $(\vec{a}-\vec{c}) \times \vec{b}=-18 \hat{i}-3 \hat{j}+12 \hat{k}$ and $\vec{a} \cdot \vec{c}=3$. If $\vec{b} \times \vec{c}=\vec{d}$,then $|\vec{a} \cdot \vec{d}|$ is equal to:

The number of unit vectors perpendicular to $\overline{a}=(1,1,0)$ and $\overline{b}=(0,1,1)$ is

Let $\vec{a}$ and $\vec{b}$ be the vectors along the diagonals of a parallelogram having area $2 \sqrt{2}$. Let the angle between $\vec{a}$ and $\vec{b}$ be acute. Given $|\vec{a}|=1$ and $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$. If $\vec{c}=2 \sqrt{2}(\vec{a} \times \vec{b})-2 \vec{b}$,then find the angle between $\vec{b}$ and $\vec{c}$.