If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0}$,then $|\vec{a} \times \vec{c}|$ is equal to

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{c} = \hat{j} - \hat{k}$ and a vector $\vec{b}$ be such that $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 3$. Then $|\vec{b}|$ equals?

Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is :-

If the vectors $\vec{AB} = p \hat{i} + q \hat{j} + r \hat{k}$,$\vec{AC} = s \hat{i} + 3 \hat{j} + 4 \hat{k}$,and $\vec{CB} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ form a $\triangle ABC$,then the values of $p, q, r$ and $s$ such that the area of that $\triangle ABC$ is $5 \sqrt{6}$ are:

If $\hat{a}$ is a unit vector such that $(\bar{x}-\hat{a}) \cdot (\bar{x}+\hat{a}) = 8$,then $|\bar{x}| = $

If $\bar{a} = \hat{i} + \hat{j}$ and $\bar{b} = 2\hat{i} - \hat{k}$,then the point of intersection of the lines $\bar{r} \times \bar{a} = \bar{b} \times \bar{a}$ and $\bar{r} \times \bar{b} = \bar{a} \times \bar{b}$ is