jmc

We first unpack the statement $x<-4$ or $|x-25|\le 1.$ The inequality $|x-25|\le 1$ is equivalent to $-1\le x-25\le 1,$ which is in turn equivalent to $24\le x\le 26.$ Therefore, we have either $x<-4$ or $24\le x\le 26,$ so the solution set for $x$ is $$(-\infty ,-4)\cup [24,26].$$The sign of the expression $\frac{(x-a)(x-b)}{x-c}$ changes at $x=a,$ $x=b,$ and $x=c,$ which means that $a,$ $b,$ and $c$ must be the numbers $-4,$ $24,$ and $26,$ in some order. Furthermore, since $24$ and $26$ are endpoints of a closed interval (that is, they are included in the solution set), it must be the case that $a$ and $b$ are $24$ and $26$ in some order, because the inequality is true when $x=a$ or $x=b,$ but is not true when $x=c$ (since that would make the denominator zero). Since $a<b,$ we have $a=24$ and $b=26,$ and then $c=-4.$

In conclusion, the given inequality must be $$\frac{(x-24)(x-26)}{x+4}\le 0.$$To check that the solution to this inequality is $(-\infty ,-4)\cup [24,26],$ we can build a sign table, where $f(x)$ is the expression on the left-hand side: \begin{array}{c|ccc|c} &$x-24$ &$x-26$ &$x+4$ &$f(x)$ \\ \hline$x<-4$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]$-4<x<24$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]$24<x<26$ &$+%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$x>26$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}This shows that $f(x)<0$ when $x\in (-\infty ,-4)\cup (24,26),$ and since $f(x)=0$ for $x\in \{24,26\},$ we indeed have the solution set $$x\in (-\infty ,-4)\cup [24,26].$$Thus, $a+2b+3c=24+2(26)+3(-4)=\boxed{64}.$