$A$ tetrahedron has vertices $P(1, 2, 1)$,$Q(2, 1, 3)$,$R(-1, 1, 2)$ and $O(0, 0, 0)$. The angle between the faces $OPQ$ and $PQR$ is

$\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$, $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}+\hat{j}-\hat{k}$ are three vectors. If $\vec{d}$ is a normal to the plane of $\vec{a}$ and $\vec{b}$ and $\vec{d} \cdot \vec{c}=2$, then $|\vec{d}|=$

If $\overline{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\overline{OB} = 4\hat{i} + \hat{k}$ are the position vectors of the points $A$ and $B$, then the position vector of a point on the line passing through $B$ and parallel to the vector $\overline{OA} \times \overline{OB}$ which is at a distance of $\sqrt{189}$ units from $B$ is

Let $\vec{a}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+3 \hat{k}$. If $\vec{b}$ is a vector such that $\vec{a}=\vec{b} \times \vec{c}$ and $|\vec{b}|^2=50$,then $|72-| \vec{b}+\vec{c}|^2 |$ is equal to $..........$.

If $\vec{a}=\hat{i}-\hat{j}+\hat{k}$,$\vec{b}=\hat{i}+\hat{j}-2 \hat{k}$,$\vec{c}=2 \hat{i}-3 \hat{j}-\hat{k}$,and $\vec{d}=2 \hat{i}+\hat{j}+\hat{k}$ are four vectors,then find the value of $(\vec{a} \times \vec{c}) \times(\vec{b} \times \vec{d})$.

$a \times (b \times c)$ is coplanar with