Find the coordinates of the point which divides the line segment joining the points $(-2, 3, 5)$ and $(1, -4, 6)$ in the ratio $2:3$ externally.

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

Using the section formula,show that the points $A(2, -3, 4)$,$B(-1, 2, 1)$,and $C(0, \frac{1}{3}, 2)$ are collinear.

If $A(4,3,5)$,$B(0,-2,2)$,and $C(3,2,1)$ are three points,then the coordinates of the point $D$ where the bisector of $\angle BAC$ meets the side $BC$ are:

If $m:n$ is the ratio in which the point $\left(\frac{8}{5}, -\frac{1}{5}, \frac{8}{5}\right)$ divides the line segment joining the points $(2, p, 2)$ and $(p, -2, p)$, where $p$ is an integer, then $\frac{3m+n}{3n} =$

The point dividing the line segment joining the points $(1, 2, 3)$ and $(3, -5, 6)$ in the ratio $3: -5$ is