If $(2,-1,3)$ is the foot of the perpendicular drawn from the origin $(0,0,0)$ to a plane, then the equation of that plane is

If $A(2,-1,1), B(2,5,1)$ and $C(0,-2,3)$ are the vertices of a triangle. If $D$ is the point of intersection of the side $B C$ and the internal angular bisector of angle $A$, then $A D=$

A plane $\pi$ given by $a x+b y+11 z+d=0$ is perpendicular to the planes $2 x-3 y+z=4$, $3 x+y-z=5$ and the perpendicular distance from the origin to the plane $\pi$ is $\sqrt{6}$ units. If all the intercepts made by the plane $\pi$ on the coordinate axes are positive, then $d=$

For a positive real number $p$, if the perpendicular distance from a point $-\hat{\mathbf{i}}+p \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ to the plane $\mathbf{r} \cdot(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})=7$ is 6 units, then $p=$