Note: I revisited this problem after I can mostly understand modular arithmetic. Since I was not too satisfied with those answers I decided to try to answer question once again with my own understanding of probability and modular arithmetic (you may see duplicate questions) and I want to know if my answer is correct:
Pick $a$, $b$, $c$ randomly without any conditions from the set $(1, 2, 3,..., 2019)$ What is the probability for $abc + bc + c$ to be divisible by 3?
Suppose $abc + bc + c$ is divisible by 3, then $abc+bc+c = c(ab+b+1)$.
Case 1: $c$ is divisible by $3$. If $c$ is divisible by $3$ then there are $673$ integers from the set that satisfies $c$. Since c is a multiple of 3 then $a$ and $b$ can be any integers from the set. In conclusion, $$P(c) = \frac{1}3, P(a) = P(b) = 1$$
and the probability for case 1 is $\frac{1}3 \cdot 1 \cdot 1 = \frac{1}3$
Case 2: $ab + b+ 1$ is divisible by $3$. Factoring $ab + b + 1 = b(a+1)+1$ Then we can utilize modular arithmetic for this case. $$b(a+1)+1\equiv 0 \pmod3$$ $$b(a+1)\equiv 2\pmod3$$ This is the part where I will use the fact that if $p$ is prime, and $xy \equiv p \pmod n$ then $x \equiv 1\pmod n$ and $y \equiv p \pmod n$ or vice versa. Then we can divide the problem to subcases.
Subcase 1: $$b \equiv 2\pmod 3 \ {and} $$ $$a+1 \equiv 1\pmod 3 \rightarrow a \equiv 0 \pmod3$$
For $b \equiv 2\pmod 3$ there are $673$ integers that satisfies $b$ ($\frac{1}3$ of the set), and for $a \equiv 0 \pmod3$ there are also $673$ integers that satisfies $a$ ($\frac{1}3$ of the set). Note that $c$ can be any integers from the set. Then the probability for subcase 1 is $\frac{1}3 \cdot \frac{1}3 \cdot 1 = \frac{1}9$
Subcase 2: $$b \equiv 1\pmod 3 \ {and} $$ $$a+1 \equiv 2\pmod 3 \rightarrow a \equiv 1 \pmod3$$
For $b \equiv 1\pmod 3$ there are $673$ integers that satisfies $b$ ($\frac{1}3$ of the set), and for $a \equiv 1 \pmod3$ there are also $673$ integers that satisfies $a$ ($\frac{1}3$ of the set). Note that $c$ can be any integers from the set. Then the probability for subcase 2 is $\frac{1}3 \cdot \frac{1}3 \cdot 1 = \frac{1}9$
Then, the total probability will be $\frac{1}3 + \frac{1}9 +\frac{1}9 = \frac{5}9$. My question is, is this the correct answer?
Note: I found where my mistake was. My mistake was, in the second case, I assumed $c$ could be anything (including multiples of 3). But assuming $c$ is a multiple of 3 automatically falls into the first case. Hence, I should've assumed $c$ is not a multiple of 3 in the second case. Then excluding 3, I should multiply $\frac{2}3$ to $\frac{2}9$
Since 2019 is divisible by 3 (other than that, it's random), we can instead solve it just for 1,2,3. If $c$ is divisible by $3$, it must be true.
If $3\nmid c$, then we need $ab+b+1$ to be divisible by 3, or $ab+b$ to be 2 mod 3. There are two solutions, modulo 3.
The answer is $\frac 13 + \frac 23 \cdot \frac 29 = \frac{13}{27}$.
By the way, the only thing you did wrong was overcount the cases where both $ab+b+1$ and $c$ are divisible by 3.