Define a relation $\sim$ on the set $\{1,2,3,4,5,6,7,8,9,10\}$ by $a \sim b$ if $a - 2b$ is divisible by $3$. Then:

Given relation:
$a \sim b$ if $a - 2b$ is divisible by $3$

This means
$a - 2b \equiv 0 \pmod 3$

So,
$a \equiv 2b \pmod 3$

Checking reflexive:
For reflexive relation, $a \sim a$ must be true for every $a$.

Now,
$a - 2a = -a$

For all values of $a$, $-a$ is not divisible by $3$.
So, relation is not reflexive.

Checking symmetric:
If $a \sim b$, then
$a \equiv 2b \pmod 3$

Multiplying both sides by $2$,
$2a \equiv 4b \pmod 3$

Since $4 \equiv 1 \pmod 3$,
$2a \equiv b \pmod 3$

So,
$b - 2a \equiv 0 \pmod 3$

Hence, $b \sim a$.
So, relation is symmetric.

Checking transitive:
Let $a \sim b$ and $b \sim c$.

Then,
$a \equiv 2b \pmod 3$

and
$b \equiv 2c \pmod 3$

So,
$a \equiv 2(2c) \pmod 3$

$a \equiv 4c \pmod 3$

Since $4 \equiv 1 \pmod 3$,
$a \equiv c \pmod 3$

But for $a \sim c$, we need
$a \equiv 2c \pmod 3$

This is not always true.
So, relation is not transitive.

Therefore, $\sim$ is symmetric but neither transitive nor reflexive.