A bisector of the angle between the normals of the planes $4 x+3 y=5$ and $x+2 y+2 z=4$ is along the vector

If $A(1,2,3), B(2,-3,1), C(3,2,-1)$ are three vertices of a tetrahedron $A B C D$ and $G\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right)$ is its centroid, then the point which divides $G D$ in the ratio $1: 2$ is

Let $D$ be the foot of the perpendicular drawn from the point $A(2,0,3)$ to the line joining the points $B(0,4,1)$ and $C(-2,0,4)$. Then, the ratio in which $D$ divides $B C$ is

Let $6 x-3 y+2 z-6=0$ be the given plane. If $a, b$ and $c$ are the intercepts made by the plane on $X, Y$ and $Z$-axes, respectively; $l, m$ and $n$ are the direction cosines of a normal drawn to the plane and $p$ is the perpendicular distance from the origin to the plane, then $|a l+b m+c n|=$