If $A(3,2,-1), B(-2,2,-3)$ and $D(-2,5,-4)$ are the vertices of a parallelogram,then the area of the parallelogram is

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:

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 $\vec{c}$.

The diagonals of a parallelogram are $\vec{d_1} = \hat{j} + \hat{k}$ and $\vec{d_2} = \hat{i} + \hat{j}$. The area of the parallelogram is . . . . . . sq. units.

If $\bar{a}=\hat{j}-\hat{k}$ and $\bar{c}=\hat{i}-\hat{j}-\hat{k}$,then the vector $\bar{b}$ satisfying $\bar{a} \times \bar{b}+\bar{c}=\vec{0}$ and $\bar{a} \cdot \bar{b}=3$ is

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$,$\overrightarrow{b}=\hat{i}+\hat{j}$,$\overrightarrow{c}=\hat{i}$ and $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$,then $\lambda+\mu$ is equal to: