The points A(3, 2, –4), B(5, 4, –6) and C(9, 8, –10) form

The given points are $A\left(3,2,-4\right),B\left(5,4,-6\right),C\left(9,8,-10\right)$

The lengths of the line segments $AB,BC,CA$ are as below

Hence, 

$\begin{array}{rcl}AB& =& \sqrt{{\left(5-3\right)}^2+{\left(4-2\right)}^2+{\left(-6+4\right)}^2}\\ & =& \sqrt{4+4+4}\\ & =& \sqrt{12}\\ & =& 2\sqrt{3}\end{array}$

$\begin{array}{rcl}BC& =& \sqrt{{\left(9-5\right)}^2+{\left(8-4\right)}^2+{\left(-10+6\right)}^2}\\ & =& \sqrt{16+16+16}\\ & =& \sqrt{48}\\ & =& 4\sqrt{3}\end{array}$

$\begin{array}{rcl}CA& =& \sqrt{{\left(9-3\right)}^2+{\left(8-2\right)}^2+{\left(-10+4\right)}^2}\\ & =& \sqrt{36+36+36}\\ & =& 6\sqrt{3}\end{array}$

Observing that $AB+BC=CA$

Therefore, the given points $A,B,C$ are collinear.