(In reply to I think I got it.
But there is a much simpler solution that covers the more general case where each of a, b and c are any nonzero real numbers. (The specific reason of stating that each of a, b and c is positive is to keep the denominator of each of the three expressions nonzero.)
Assume that (ay-bx)/c = (cx-az)/b = (bz-cy)/a =p(say)
Then arrive at:
A quadratic expression involving a, b and c multiplied by something is equal to zero and, everything else will just fall in place.
The quadratic expression and that something are much more simpler than envisaged.
Edited on November 25, 2012, 3:36 am