Vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are such that $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$. $P_1$ and $P_2$ are two planes determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}, \vec{d}$ respectively. Then the angle between the planes $P_1$ and $P_2$ is

$\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors and $\vec{a}$ is a unit vector such that $\cos (\vec{a}, \vec{b} \times \vec{c})=\sqrt{\frac{2}{3}}$. Then $|\vec{a} \times(\vec{b} \times \vec{c})|=$

If $|\vec{a}| = 4$,$|\vec{b}| = 2$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$,then $|\vec{a} \times \vec{b}|^2 = \dots$

Let $\vec{a}=2 \hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}-5 \hat{k}$ be two vectors and $\vec{r}$ be a vector in the plane of $\vec{a}$ and $\vec{b}$. If $\vec{r}$ is orthogonal to the vector $5 \hat{i}-2 \hat{j}+3 \hat{k}$ and the magnitude of $\vec{r}$ is $\sqrt{94}$,then $|\vec{r} \cdot \vec{b}|=$

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

Find a unit vector perpendicular to each of the vectors $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b}),$ where $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}+3\hat{k}.$