jmc

In the first equation, adding $(a-13)x$ to both sides gives us $ax+by+cz=(a-13)x$. Solving for $x$, we have $$x=\frac{ax+by+cz}{a-13}.$$Since $a\ne 13$ and $x\ne 0$, both sides of the equation are non-zero. Similarly from the 2nd and 3rd equation, $$y=\frac{ax+by+cz}{b-23}$$and $$z=\frac{ax+by+cz}{c-42}.$$Then we know that $$\begin{array}{rl}ax+by+cz& =a\cdot \frac{ax+by+cz}{a-13}+b\cdot \frac{ax+by+cz}{b-23}+c\cdot \frac{ax+by+cz}{c-42}\\ & =(ax+by+cz)\left(\frac{a}{a-13}+\frac{b}{b-23}+\frac{c}{c-42}\right).\end{array}$$If $ax+by+cz=0$, then $x=\frac{ax+by+cz}{a-13}=0$. But we know $x\ne 0$. Hence, $ax+by+cz\ne 0$. Thus, $$\frac{a}{a-13}+\frac{b}{b-23}+\frac{c}{c-42}=\boxed{1}.$$