I've seen relations defined as functions between sets and as sets of ordered sets; however, I've never seen a relation defined as $3\mid(a+2b)$. What does this mean?
--Edit--
I'll try and express my reasoning. Presuming I understand the relation as he expressed it, I presume it is symmetrical since if a=b then 3a is divisible by 3, and thus a multiple of it. For symmetry, it would not seem to matter whether one assigns value x to a and value y to b or vice versa, in either case the sum is the same, so they relate if 3 divides that sum. For transitivity, there is no third variable under consideration so I suppose it is vacuously true that they are transitive.
The relation means that $(a, b)\in R$, or $a \sim b$ ($a$ is related to $b$) if and only if $3$ divides $a + 2b$. So, for example, $(1, 1) \in R$, since $1 + 2(1) = 3$, which is certainly divisible by $3$.
You need to show whether the relation is reflexive, symmetric, and/or transitive.
E.g. Reflexivity: $\forall x \in \mathbb Z$, is $(x, x) \in R$? Well $x + 2x = 3x$, which is divisible by $3$. So the relation is reflexive.
Symmetry: Is it the case that for all $x, y \in \mathbb Z$, if $(x, y) \in R$, then $(y,x) \in R$? Suppose $(x, y) \in R$. Then $3\mid x + 2y$. This means that $x+2y = 3k$ for some integer $k$. Certainly, we also know that $3\mid 3(x+y)$. Now, $y + 2x = 3(x+y) - (x +2y)$ must be divisible by $3$ since $3(x+y)$ and $x+2y=(3k)$ is divisible by $3$. Hence, if $(x, y) \in R,$ then so too $(y, x) \in R$.
Transitivity: Is it the case that So for $x, y, z \in \mathbb Z$, IF $(x, y) \in R$ and $(y, z)\in R$, it follows that $(x, z) \in R$? I'll let you work at tackling this. Assume $(x, y)\in R,$ and $(y, z)\in R$. Now, unpack what this means, and from this assumption, check whether it then follows that $3\mid x + 2z$, i.e., if it then follows that $(x, z)\in R$.