If the area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is $20$ square units,then the area of the parallelogram having $3 \bar{a} + \bar{b}$ and $2 \bar{a} + 3 \bar{b}$ as two adjacent sides in square units is

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. If $\overline{c}$ is a vector such that $\overline{a} \cdot \overline{c}=|\overline{c}|$,$|\overline{c}-\overline{a}|=2 \sqrt{2}$,and the angle between $\overline{a} \times \overline{b}$ and $\overline{c}$ is $\frac{2 \pi}{3}$,then $|(\overline{a} \times \overline{b}) \times \overline{c}|=$

Find $\lambda$ and $\mu$ if $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}$.

$(a - b) \times (a + b) = $

Let $\lambda \in R$,$\vec{a} = \lambda \hat{i} + 2 \hat{j} - 3 \hat{k}$,and $\vec{b} = \hat{i} - \lambda \hat{j} + 2 \hat{k}$. If $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$,then $|\lambda(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|^2$ is equal to

If $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$,$\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$,$|\overrightarrow{a}|=|\overrightarrow{b}|=2$,then $|\overrightarrow{u} \times \overrightarrow{v}|$ is equal to