If the position vectors of $P$ and $Q$ are $\hat{i}+2 \hat{j}-7 \hat{k}$ and $5 \hat{i}-3 \hat{j}+4 \hat{k}$ respectively, then the cosine of the angle between $\overrightarrow{PQ}$ and $z$-axis is

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ are the position vectors of the vertices $A$,$B$ and $C$ respectively of triangle $ABC$,then the position vector of the point in which the bisector of $\angle B$ meets $CA$ is:

Let $\overrightarrow{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\overrightarrow{b} = 7\hat{i} + \hat{j} - 6\hat{k}$. If $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{r} \times \overrightarrow{b}$ and $\overrightarrow{r} \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$,then $\overrightarrow{r} \cdot (2\hat{i} - 3\hat{j} + \hat{k})$ is equal to:

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $2 \vec{a}+3 \vec{b}+4 \vec{c}=\vec{0}$. Then $|\vec{b} \times \vec{c}|=$

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

If in a right-angled triangle $ABC$,the hypotenuse $|\overrightarrow{AB}| = p$,then $\overrightarrow{AB} \cdot \overrightarrow{AC} + \overrightarrow{BC} \cdot \overrightarrow{BA} + \overrightarrow{CA} \cdot \overrightarrow{CB} = $