If the area of the parallelogram with $a$ and $b$ as two adjacent sides is $15$ sq units,then the area of the parallelogram having $3a+2b$ and $a+3b$ as two adjacent sides in sq units is

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\pi / 3$,then $\vec{a}$ is equal to

If $\bar{a}=2 \hat{i}+3 \hat{j}-\hat{k}, \bar{b}=-\hat{i}+2 \hat{j}-4 \hat{k}$ and $\bar{c}=\hat{i}+\hat{j}-2 \hat{k}$,then $(\bar{a} \times \bar{b}) \cdot(\bar{a} \times \bar{c})=$

The vectors are $\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\bar{b}=\hat{i}+\hat{j}$. If $\bar{c}$ is a vector such that $\bar{a} \cdot \bar{c}=|\bar{c}|$ and $|\bar{c}-\bar{a}|=2 \sqrt{2}$,and the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$ is $\frac{\pi}{4}$,then find the value of $|(\bar{a} \times \bar{b}) \times \bar{c}|$.

For $\vec{a}$ and $\vec{b}$,$|\vec{a}|=3$,$|\vec{b}|=\frac{\sqrt{2}}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector,then the angle between $\vec{a}$ and $\vec{b}$ is . . . . . . .

$A$ tetrahedron has vertices $O(0,0,0)$,$A(1,2,1)$,$B(2,1,3)$,and $C(-1,1,2)$. The angle between the faces $OAB$ and $ABC$ is: