Let $|\vec{a}|=2, |\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{4}$. Then $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2$ is equal to

If $\vec{r}$ is a unit vector satisfying $\vec{r} \times \vec{a}=\vec{b}$, $|\vec{a}|=2$ and $|\vec{b}|=\sqrt{3}$, then one such $\vec{r}=$

For any two vectors $a$ and $b$,if $a \times b = 0$,then

If $a$ and $b$ are unit vectors such that $a \times b$ is also a unit vector,then the angle between $a$ and $b$ is

If non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular to each other,then the solution of the equation $\vec{r} \times \vec{a} = \vec{b}$ is:

$\vec{a}, \vec{b}, \vec{c}$ are three vectors each having $\sqrt{2}$ magnitude such that $(\vec{a}, \vec{b})=(\vec{b}, \vec{c})=(\vec{c}, \vec{a})=\frac{\pi}{3}$. If $\vec{x}=\vec{a} \times(\vec{b} \times \vec{c})$ and $\vec{y}=\vec{b} \times(\vec{c} \times \vec{a})$, then