Factoring- Pre Algebra Homework question

I'm studying pre-algebra with an App called Brilliant. I'm in a lesson called "Factoring Sums with Variables". I got most of the exercises right, but there's one that I don't understand. To begin with, I don't even understand the question :( The bold part of the text is what I don't understand. Where did they get the $2ac$ from?

Question: Which of these equations does not represent a possible solution to the equation?

$ 6a^2bc+18a^2b^2c-24abc^2=0$?

1. $3ab+9ab^2-12bc=0$
2. $2a^2bc+3a^2b^2c-4abc^2=0$
3. $2a^2+6a^2b-8ac=0$
4. $a + 3ab-4c=0$

It gave me the following hint:

"First, we have $3ab+9ab^2-12bc.$ If we multiply by $2ac$ and distribute, we get:

$2ac(3ab+9ab^2-12bc)$
$=6a^2bc+18a^2b^2c-24abc^2$

So the equation is equivalent to:

$2ac(3ab+9ab^2-12bc)=0$
and by the zero-product property, we know that possible solutions are when $2ac=0$ or $3ab+9ab^2-12bc=0$

Solution 1:

Your equation can be written as $$ 6abc(a+3ab-4c)=0. $$ In a field with $6\neq 0$, either $abc=0$, or $a+3b-4c=0$, which is equation number $4$. So this equation definitely can be a solution. To see that equation $2$ is false in general, take $a=2$, $b=5$ and $c=8$. Then the original equation is true, but not equation $2$. Actually, all others can be solutions.