Using the section formula,prove that the three points $A(-4, 6, 10)$,$B(2, 4, 6)$,and $C(14, 0, -2)$ are collinear.

(N/A) Let the points be $A(-4, 6, 10)$,$B(2, 4, 6)$,and $C(14, 0, -2)$.
Assume that a point $P$ divides the line segment $AB$ in the ratio $k:1$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left( \frac{2k - 4}{k + 1}, \frac{4k + 6}{k + 1}, \frac{6k + 10}{k + 1} \right)$
For the points to be collinear,point $C$ must coincide with point $P$ for some value of $k$.
Equating the $x$-coordinates: $\frac{2k - 4}{k + 1} = 14$
$2k - 4 = 14k + 14 \implies -12k = 18 \implies k = -\frac{18}{12} = -\frac{3}{2}$
Now,check the $y$-coordinate for $k = -\frac{3}{2}$:
$\frac{4(-\frac{3}{2}) + 6}{-\frac{3}{2} + 1} = \frac{-6 + 6}{-\frac{1}{2}} = 0$
Now,check the $z$-coordinate for $k = -\frac{3}{2}$:
$\frac{6(-\frac{3}{2}) + 10}{-\frac{3}{2} + 1} = \frac{-9 + 10}{-\frac{1}{2}} = \frac{1}{-\frac{1}{2}} = -2$
Since the coordinates match point $C(14, 0, -2)$,the point $C$ lies on the line passing through $A$ and $B$.
Thus,the points $A$,$B$,and $C$ are collinear.