If $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$, $|\vec{a}|=3$, $|\vec{b}|=5$, and $|\vec{c}|=7$, then the angle between $\vec{a}$ and $\vec{b}$ is

$7 \bar{i}-4 \bar{j}+7 \bar{k}, \bar{i}-6 \bar{j}+10 \bar{k}, -\bar{i}-3 \bar{j}+4 \bar{k}, 5 \bar{i}-\bar{j}+\bar{k}$ are the position vectors of the points $A, B, C, D$ respectively. If $p \bar{i}+q \bar{j}+r \bar{k}$ is the position vector of the point of intersection of the diagonals of the quadrilateral $ABCD$, then $p+q+r=$

If $\bar{a}$ and $\bar{b}$ are vectors such that $|\bar{a}+\bar{b}|=\sqrt{29}$ and $\bar{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \bar{b}$,then a possible value of $(\bar{a}+\bar{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is

The angle between two consecutive sides $\vec{a}$ and $\vec{b}$ of a parallelogram is $\frac{\pi}{6}$. Given $\vec{a} = (2, -2, 1)$ and $\vec{b} = 2|\vec{a}|$,the area of the parallelogram is . . . . . . sq. units.

If the vectors $\vec{a} = \hat{i} - 2x\hat{j} - 3y\hat{k}$ and $\vec{b} = \hat{i} + 3x\hat{j} + 2y\hat{k}$ are orthogonal to each other, then the locus of the point $(x, y)$ is

If $\theta$ is the angle between the vectors $4 \hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+3 \hat{j}-2 \hat{k}$, then $\sin 2 \theta=$