For any vector $\vec{a} \in \mathbb{R}^3$,$|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = $ . . . . . . .

If $|a| = 4$,$|b| = 2$ and the angle between $a$ and $b$ is $\pi/6$,then find $|a \times b|^{2}$.

If $\bar{a}, \bar{b}, \bar{c}$ are three coplanar vectors such that $|\bar{a}|=1, |\bar{b}|=2$,$\bar{b} \cdot \bar{c}=8$,and the angle between $\bar{b}$ and $\bar{c}$ is $45^{\circ}$,then $|\bar{a} \times (\bar{b} \times \bar{c})|=$

Let $\overrightarrow{a}=3 \hat{i}+2 \hat{j}+\hat{k}$,$\overrightarrow{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\overrightarrow{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$ and $(\overrightarrow{a}-\overrightarrow{b}+\hat{i}) \cdot \overrightarrow{c}=-3$. Then $|\overrightarrow{c}|^2$ is equal to . . . . . . .

If $a$ and $b$ are unit vectors,then the vector $(a+b) \times (a \times b)$ is parallel to the vector

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 $\vec{c}$.