If $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$, $\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$ and $\vec{c}$ is a vector such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3$, then $\vec{a} \cdot(\vec{c} \times \vec{b}-\vec{b}-\vec{c})=$

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

If $\bar{a}=\hat{j}-\hat{k}$ and $\bar{c}=\hat{i}-\hat{j}-\hat{k}$,then the vector $\bar{b}$ satisfying $\bar{a} \times \bar{b}+\bar{c}=\vec{0}$ and $\bar{a} \cdot \bar{b}=3$ is

Vectors $\vec{p}=a \hat{i}+b \hat{j}+c \hat{k}$, $\vec{q}=d \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}$ form a triangle $ABC$ such that $\vec{p}=\vec{q}+\vec{r}$. If the area of $\triangle ABC$ is $5 \sqrt{6}$ sq. units, then the sum of the absolute values of $a, b, c$ is

If $\vec{a}_1$ is the component of vector $\vec{a}$ along the direction of vector $\vec{b}$,and $\vec{a}_2$ is the component of $\vec{a}$ perpendicular to $\vec{b}$,then $\vec{a}_1 \times \vec{a}_2 = \dots$

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 . . . . . . .