If $a, b, c, d$ are coplanar vectors,then $(a \times b) \times (c \times d)$ is equal to

If the area of the parallelogram with $\bar{a}$ and $\bar{b}$ as two adjacent sides is $15$ square units,then the area (in square units) of the parallelogram,having $3 \bar{a} + 2 \bar{b}$ and $\bar{a} + 3 \bar{b}$ as two adjacent sides,is

Let $\overrightarrow{a} = \alpha \hat{i} + 3 \hat{j} - \hat{k}$,$\overrightarrow{b} = 3 \hat{i} - \beta \hat{j} + 4 \hat{k}$ and $\overrightarrow{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ where $\alpha, \beta \in \mathbb{R}$,be three vectors. If the projection of $\overrightarrow{a}$ on $\overrightarrow{c}$ is $\frac{10}{3}$ and $\overrightarrow{b} \times \overrightarrow{c} = -6 \hat{i} + 10 \hat{j} + 7 \hat{k}$,then the value of $\alpha + \beta$ is equal to:

Let $\bar{a}=\alpha \hat{i}+3 \hat{j}-\hat{k}$,$\bar{b}=3 \hat{i}-\beta \hat{j}+4 \hat{k}$ and $\overline{c}=\hat{i}+2 \hat{j}-2 \hat{k}$,where $\alpha, \beta \in R$,be three vectors. If the projection of $\overline{a}$ on $\overline{c}$ is $\frac{10}{3}$ and $\bar{b} \times \bar{c}=-6 \hat{i}+10 \hat{j}+7 \hat{k}$,then the value of $2 \alpha+\beta$ is

The vector $x\hat{i} + y\hat{j} + z\hat{k}$ makes an acute angle $\cot^{-1} \sqrt{2}$ with the plane containing the vectors $(2, 3, -1)$ and $(1, -1, 2)$. Then,

For any three vectors $\vec{a}, \vec{b}, \vec{c}$,the value of $\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b})$ is equal to: