Thanks Melody, it's nice to know that someone is actually interested.
Often that doesn't appear to be the case.
In answer to your question, I arrived at the solution algebraically.
The equations can be written in the form
\(\displaystyle 2X+Y+Z\equiv 0 \dots\dots(1)\\ X+2Y+Z\equiv R\dots\dots(2)\\ X+Y+2Z\equiv S\dots\dots(3)\)
where each one is mod 13.
\(\displaystyle (1) - 2(3) : -Y-3Z \equiv -2S\dots\dots(4)\\ (2) - (3) :Y - Z \equiv R-S\dots\dots(5)\\ (4)+(5) : -4Z\equiv R-3S\dots\dots(6).\)
Returning to the original unknowns,
\(\displaystyle -4ca \equiv 6abc-24abc \equiv -18abc,\)
from which
\(\displaystyle9b=2+13k\quad \text{ where k is an integer}.\)
The smallest value of k for which b will be an integer is k = 4, leading to 9b = 54, b = 6.
The next suitable value for k puts b out of the permitted range.
The remaining unknowns (a = 3 and c = 9) can then be found by back substitution, (6) into (5) and then into (2) or (3).
Leave that to you.
