Find the coordinates of the points which trisect the line segment joining the points $P(4, 2, -6)$ and $Q(10, -16, 6)$.

In the triangle with vertices $A(3,2,0)$,$B(5,3,2)$,and $C(-9,6,-3)$,the bisector of $\angle BAC$ meets $BC$ at $D$. The coordinates of $D$ are

The ratio in which $B\left(\frac{33}{5}, \frac{28}{5}, \frac{38}{5}\right)$ divides the line segment joining $A(3, 2, 4)$ and $C(9, 8, 10)$ is

If the points $A(9, 8, -10)$,$B(3, 2, -4)$ and $C(5, 4, -6)$ are collinear,then the point $C$ divides the line segment $AB$ in the ratio:

Let $A(4,3,5), B(1,-2,1), C(3,2,1)$ be the vertices of a triangle $ABC$. If the internal bisector of $\angle BAC$ meets the side $BC$ at $D$, then $CD=$

Let $A(3, 2, -4)$ and $B(9, 8, -10)$ be two points. Let $P_1$ divide $AB$ in the ratio $1:2$ and $P_2$ divide $AB$ in the ratio $2:1$. If the point $P(\alpha, \beta, \gamma)$ divides $P_1 P_2$ in the ratio $1:1$, then $\alpha + 2\beta + 2\gamma =$