Let's simplify the following fractional expression:

$$
\frac{6 a^{2}-2 b^{2}+6 c^{2}+a b-13 a c-4 b c-18 a-5 b+17 c+12}{4 a^{2}-b^{2}+9 c^{2}-12 a c-16 a+24 c+16}
$$

(It is more efficient to start with the factorization of the denominator.)
The denominator can be factored as the difference of two squares - indeed, it contains $\left(-b^{2}\right)$ and $b$ does not appear in any other term - and the other 6 terms can be written as perfect squares. Among these perfect squares are: $4 a^{2}, 9 c^{2}$, and 16, whose square roots are respectively: $2 a$ and $-2 a$, similarly $\pm 3 c$, and $\pm 4$; and each of the double products formed from these can be found, and from their signs we see that the base of the square should have $|3 c|$ and $|4|$ with the same sign, while $|2 a|$ should have the opposite sign of those. According to this, the denominator:

$$
(2 a-3 c-4)^{2}-b^{2}=(2 a-3 c-4+b)(2 a-3 c-4-b)
$$

We expect that the fraction will be simplifiable by one of these two factors.

Let's denote the expected common factor of the numerator and the denominator as follows:

$$
A=2 a-3 c-4+k b
$$

where either $k=1$ or $k=-1$, and let's look for that factor $B$ which, when multiplied by this, gives the numerator. The terms $3a$ are offered by the division $\left(6 a^{2}\right):(2 a)$, similarly $(-2 c)$ from $6 c^{2}:(-3 c)$, $(-3)$ from $(+12):(-4)$, and $(-2 k b)$ from $\left(-2 b^{2}\right):(k b)$, since in both cases $k^{2}=1$. So we try the factor

$$
B=3 a-2 c-3-2 k b
$$

Subtracting $A \cdot B$ from the numerator, the remainder can be shaped as:

$$
(1+k)(a b-4 b c-5 b)
$$

This is 0 when $k=-1$, so in this case of $k$, the numerator is $A \cdot B$, and the simplifying factor is $2 a-3 c-4-b$. Thus,

$$
T=\frac{3 a-2 c-3+2 b}{2 a-3 c-4+b}
$$

and no further simplification is expected.

Of course, the simplification is only valid for such triples of numbers $(a, b, c)$ for which $2 a-3 c-4-b \neq 0$; otherwise, the original expression is undefined. Furthermore, only for such that the remaining denominator is not 0.