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:

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

The point $\left( \frac{1}{2}, -\frac{13}{4} \right)$ divides the line segment joining the points $(3, -5)$ and $(-7, 2)$ in the ratio of:

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

$A(3, 2, -1), B(4, 1, 0), C(2, 1, 4)$ are the vertices of a triangle $ABC$. If the bisector of $\angle BAC$ intersects the side $BC$ at $D(p, q, r)$,then $\sqrt{2p + q + r} =$

The point dividing the line segment joining $(3, -2, 1)$ and $(-2, 3, 11)$ in the ratio $2:3$ is