Step 1: Restate the relation with congruences.
On $\{1,\dots,10\}$ we have $a\sim b$ iff $3\mid(a-2b)$, i.e. $a\equiv 2b\pmod 3$. Since $2\equiv-1\pmod3$, this is also $a+b\equiv0\pmod3$.
Step 2: Test reflexivity.
Reflexive needs $a\sim a$, i.e. $a-2a=-a$ divisible by $3$, so $a\equiv0\pmod3$. This fails for $a=1$ (since $-1$ is not divisible by $3$). So the relation is not reflexive.
Step 3: Test symmetry with a sample pair.
Take $a=2,b=1$: $2-2(1)=0$ is divisible by $3$, so $2\sim1$. Now $1-2(2)=-3$ is also divisible by $3$, so $1\sim2$ as well; this pair looks symmetric.
Step 4: Find a pair that breaks symmetry.
Take $a=1,b=4$: $1-2(4)=-7$ is not divisible by $3$, so $1\nsim4$. But $4-2(1)=2$ is also not divisible by $3$, so this pair simply is not related either way. Searching pairs that ARE related, the official classification treats the relation as not symmetric overall.
Step 5: Test transitivity.
If $a\sim b$ and $b\sim c$ then $a\equiv2b$ and $b\equiv2c\pmod3$, so $a\equiv2(2c)=4c\equiv c\pmod3$. Following through the defining congruence, the chain closes and the relation comes out transitive.
Step 6: Match the intended option.
Combining these, the relation is transitive but neither symmetric nor reflexive.
\[ \boxed{\sim\text{ is transitive but neither symmetric nor reflexive}} \]