How do I resolve this vector equation algebraically when there are many variables? So if we have a, b, c vectors which lie on a plane and are connected through the equation:
$m\textbf{a} + n \textbf{b} + k\textbf{c} = 0$ , where  $m, n, k, \geq 0$
The task is to show that the criteria for position vectors xa, yb, zc to be collinear is:
$\frac{m}{x} + \frac{n}{y} + \frac{k}{z} = 0$
I think for the lines to be collinear, they must be parallel to each other, but to be collinear they must probably share a same point, which I don't know if it is possible to find one given so many algebraic values, or if this is the right direction to head. One way I thought of interpreting this question is to set every value of m,n,k to be 0, which makes the equation valid directly. In which direction should I approach this problem?

Before I start with the long answer I want to comment that even though $m=n=k=0$ solves the equation, this is not the answer as the question is looking for every possible answer, not just one.
The position vectors refer to not the vectors (lines) themselves, but to the points. A set of points is collinear if they all lie in the same line. Note that this is true for any two points as you can always draw a line between two points. This is not necessarily true for three points-you will usually get a triangle as opposed to a line.
Two show two lines are the same, it is sufficient to show that they have the same direction, and they share a point. If we can show that the three lines given by your point set are the same, then all three points lie in a line, and they are collinear. These are the lines between $ax$ and $by$, $by$ and $cz$,  and $cz$ and $ay$.
Note that each pair of lines shares one end point with each other line, therefore we need only show that they have the same direction.
The direction of the lines (up to a constant) are given by $ax-by$, $ax-cz$ and $by-cz$.
For them to have the same direction, there needs to be constants $h_1, h_2$ such that
 $$ax-by=h_1(ax-cz)=h_2(by-cz)$$.
I have to run right now unfortunately, but this is a start. Try using the equation of the plane to simplify these equations.
