First, consider the domain of $\frac{1}{\ln(10-x)}$. We need $10-x > 0$ and $\ln(10-x) \neq 0$.

From $10-x > 0$, we get $x < 10$.

From $\ln(10-x) \neq 0$, we get $10-x \neq 1$, which means $x \neq 9$.

So, for the first term, the domain is $x \in (-\infty, 9) \cup (9, 10)$.

Next, consider the domain of $\sin^{-4}(\frac{x+2}{2x+3})$. Since it is raised to the power of -4, $\sin^{-4}(\frac{x+2}{2x+3}) = \frac{1}{\sin^4(\frac{x+2}{2x+3})}$.

We need $-1 \leq \frac{x+2}{2x+3} \leq 1$ and $\sin(\frac{x+2}{2x+3}) \neq 0$, which means $\frac{x+2}{2x+3} \neq 0$.

First, let's solve $-1 \leq \frac{x+2}{2x+3} \leq 1$.

Case 1: $\frac{x+2}{2x+3} \leq 1 \implies \frac{x+2}{2x+3} - 1 \leq 0 \implies \frac{x+2 - (2x+3)}{2x+3} \leq 0 \implies \frac{-x-1}{2x+3} \leq 0 \implies \frac{x+1}{2x+3} \geq 0$.

The critical points are $x = -1$ and $x = -\frac{3}{2}$. Thus, $x \in (-\infty, -\frac{3}{2}) \cup [-1, \infty)$.

Case 2: $\frac{x+2}{2x+3} \geq -1 \implies \frac{x+2}{2x+3} + 1 \geq 0 \implies \frac{x+2 + 2x+3}{2x+3} \geq 0 \implies \frac{3x+5}{2x+3} \geq 0$.

The critical points are $x = -\frac{5}{3}$ and $x = -\frac{3}{2}$. Thus, $x \in (-\infty, -\frac{5}{3}] \cup (-\frac{3}{2}, \infty)$.

Combining both cases, we have $x \in (-\infty, -\frac{5}{3}] \cup (-1, \infty)$.

Now, we need to consider $\frac{x+2}{2x+3} \neq 0$, which means $x \neq -2$.

So, the domain for the second term is $(-\infty, -\frac{5}{3}] \cup (-2, -\frac{3}{2}) \cup (- \frac{3}{2}, -1) \cup (-1, \infty)$.

Combining the domains of both terms, we have:

$x \in [(-\infty, 9) \cup (9, 10)] \cap [(-\infty, -\frac{5}{3}] \cup (-2, -\frac{3}{2}) \cup (- \frac{3}{2}, -1) \cup (-1, \infty)]$.

$x \in (-\infty, -\frac{5}{3}] \cup (-1, 9) \cup (9, 10)$.

Given the domain is $(-\infty, -a] \cup (-1, b) \cup (b, c)$, we can identify $a = \frac{5}{3}$, $b = 9$, and $c = 10$.

Finally, we need to find $b+c+3a = 9 + 10 + 3(\frac{5}{3}) = 9 + 10 + 5 = 24$.