The figure formed by the four points $i + j - k$,$2i + 3j$,$3i + 5j - 2k$,and $k - j$ is:

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$,$|\vec{a}| = 1$,$|\vec{b}| = 2$,and $|\vec{c}| = 3$,then find the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.

Let $\overrightarrow{OA}$ and $\overrightarrow{OB}$ be two sides of a triangle. The median $\overrightarrow{AM}$ is perpendicular to the angle bisector $\overrightarrow{OL}$ and $|\overrightarrow{AM}|:|\overrightarrow{OL}|=1:2$. The angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is

If $A=(1,-1,2)$,$B=(3,4,-2)$,$C=(0,3,2)$ and $D=(3,5,6)$,then the angle between the lines $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is (in $^{\circ}$)

If $7 \hat{i}-4 \hat{j}+5 \hat{k}$ is the position vector of the vertex $A$ of a tetrahedron $ABCD$ and $-\hat{i}+4 \hat{j}-3 \hat{k}$ is the position vector of the centroid of the triangle $BCD$,then the position vector of the centroid of the tetrahedron $ABCD$ is

Let $PQR$ be a triangle such that $\overrightarrow{PQ}=-2\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{PR}=a\hat{i}+b\hat{j}-4\hat{k}$,where $a, b \in \mathbb{Z}$. Let $S$ be the point on $QR$,which is equidistant from the lines $PQ$ and $PR$. If $|\overrightarrow{PR}|=9$ and $\overrightarrow{PS}=\hat{i}-7\hat{j}+2\hat{k}$,then the value of $3a-4b$ is . . . . . . .