Let $ M= \left[ {\begin{array}{ccc} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{array} } \right] $, such that $a_{1,n} + a_{2,n} + a_{3,n} = 1.$
As an example, $M$ could equal $ \left[ {\begin{array}{ccc} .3 & .1 & .8 \\ .5 & .4 & .1 \\ .2 & .5 & .1 \\ \end{array} } \right] $.
Let $ v= \left[ {\begin{array}{c} x \\ y \\ z \\ \end{array} } \right] $, where $x$, $y$, and $z$ are all positive.
The geometric sequence $v, Mv, M^2v...$ define a series of vectors. Adding those vectors to the origin gives us a series of points. Show that these points are co-planar.
I've done some reading on this and some people have mentioned Eigenvectors, etc. I am currently in precalc and so I'd rather have nothing more advanced than calculus be used. I also think this is a Markov chain so I'm also adding that to the tags, feel free to remove it if you disagree.
Given some thought, I've realized the answer is simple. Let $W = x + y + z$. Because this is a Markov chain, for all $M^nv$, $x_{M^nv}+y_{M^nv}+z_{M^nv} = W$. The equation this line fits is $x + y + z = W$. Thanks for the help everyone. I'm amazed I did not see this last night.