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

The direction cosines of a normal to the plane passing through $(4,2,3)$,$(-1,4,2)$ and $(3,2,1)$ are .....

The perpendicular distance from the origin to the plane $x + 2y - 2z + 5 = 0$ equals $.........$ units.

The equation of the plane passing through the points with position vectors $A(2 \hat{i}+6 \hat{j}-6 \hat{k})$, $B(-3 \hat{i}+10 \hat{j}-9 \hat{k})$ and $C(-5 \hat{i}-6 \hat{k})$ is

If the vector equation of the plane $\bar{r}=(2 \hat{i}+\hat{k})+\lambda \hat{i}+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ in scalar product form is given by $\bar{r} \cdot(3 \hat{j}+2 \hat{k})=\alpha$,then $\alpha=$

The distance of a point $\vec{a}$ from the plane $\vec{r} \cdot \vec{m} = q$ is given by $\frac{|\vec{a} \cdot \vec{m} - q|}{|\vec{m}|}$. If the distance of the point $\hat{i} + 2\hat{j} + 3\hat{k}$ from the plane $\vec{r} \cdot (2\hat{i} + 6\hat{j} - 9\hat{k}) = -1$ is $p$ and the distance of the origin from this plane is $q$,then $p - q =$