1. Given relation

We have a relation $R$ on the set $A\times B$, where $A=\{2,3,6,8,9,11\},\qquad B=\{1,4,5,10,15\}.$

For $(a,b),(c,d)\in A\times B$, $(a,b)R(c,d) \iff 3ad-7bc \text{ is even.}$

We must determine whether $R$ is reflexive, symmetric, transitive, etc.

2. Simplify the condition using parity

Since we only care whether $3ad-7bc$ is even, work modulo $2$.

Now, $3\equiv 1 \pmod 2, \qquad 7\equiv 1 \pmod 2.$ So $3ad-7bc \equiv ad-bc \pmod 2.$ But modulo $2$, subtraction and addition are the same, so $ad-bc \equiv ad+bc \pmod 2.$ Thus, $(a,b)R(c,d) \iff ad-bc \equiv 0 \pmod 2.$ Equivalently, $(a,b)R(c,d) \iff ad \equiv bc \pmod 2.$

So the relation depends only on the parity of $a,b,c,d$.

3. Parity structure of elements of $A$ and $B$

From $A$:

• even elements: $2,6,8$
• odd elements: $3,9,11$

From $B$:

• even elements: $4,10$
• odd elements: $1,5,15$

For any pair $(a,b)\in A\times B$, there are four parity types:

• $(E,E)$
• $(E,O)$
• $(O,E)$
• $(O,O)$

Now compute $ad \equiv bc \pmod 2$ for these types.

Let us note:

• product is even if at least one factor is even,
• product is odd only if both factors are odd.

4. Check reflexivity

For reflexivity, every $(a,b)$ must satisfy $(a,b)R(a,b).$

This means $3ab-7ba = (3-7)ab = -4ab,$ which is always even.

Hence $R$ is reflexive.

5. Check symmetry

Assume $(a,b)R(c,d)$. Then $3ad-7bc \text{ is even.}$ We need to test whether $(c,d)R(a,b)$ holds, i.e. whether $3cb-7da$ is even.

Again modulo $2$, $3ad-7bc \equiv ad-bc \equiv ad+bc \, (\bmod 2),$ and $3cb-7da \equiv cb-da \equiv cb+da \, (\bmod 2).$ But since multiplication is commutative, $cb+da = bc+ad.$ So both expressions have the same parity.

Therefore, $(a,b)R(c,d) \implies (c,d)R(a,b).$ Thus $R$ is symmetric.

6. Check transitivity

We must see whether $(a,b)R(c,d) \text{ and } (c,d)R(e,f) \implies (a,b)R(e,f).$

We will find a counterexample.

Take: $(a,b)=(2,1),\quad (c,d)=(3,4),\quad (e,f)=(2,5).$ These are all in $A\times B$ since

• $2,3\in A$, $1,4,5\in B$.

Now check each relation:

(i) $(2,1)R(3,4)$

$3\cdot 2\cdot 4 - 7\cdot 1\cdot 3 = 24-21=3,$ which is odd. So this does not work.

Let us choose a better example using parity.

We want:

• first pair related,
• second pair related,
• first and third not related.

Since the condition is $ad\equiv bc\pmod 2$, classify by parity type:

• $(E,E)$ behaves like both products always even with many types,
• $(E,O)$ and $(O,E)$ are often related,
• $(O,O)$ behaves differently.

Let us explicitly test using parity types.

For a pair $(x,y)$, define:

• $x$ parity from $A$,
• $y$ parity from $B$.

The relation condition between $(a,b)$ and $(c,d)$ is $ad \equiv bc \pmod 2.$

Now take

• $(a,b)$ of type $(E,O)$, say $(2,1)$,
• $(c,d)$ of type $(E,E)$, say $(6,4)$,
• $(e,f)$ of type $(O,E)$, say $(3,4)$.

Check:

(i) $(2,1)R(6,4)$

$3\cdot 2\cdot 4 - 7\cdot 1\cdot 6 = 24-42=-18,$ which is even. So $(2,1)R(6,4)$ is true.

(ii) $(6,4)R(3,4)$

$3\cdot 6\cdot 4 - 7\cdot 4\cdot 3 = 72-84=-12,$ which is even. So $(6,4)R(3,4)$ is true.

(iii) $(2,1)R(3,4)$

$3\cdot 2\cdot 4 - 7\cdot 1\cdot 3 = 24-21=3,$ which is odd. So $(2,1)R(3,4)$ is false.

Thus, $(2,1)R(6,4),\quad (6,4)R(3,4),\quad \text{but } (2,1)\not R(3,4).$

Hence $R$ is not transitive.

7. Conclusion

We have shown:

• $R$ is reflexive,
• $R$ is symmetric,
• $R$ is not transitive.

Therefore, the correct option is $\boxed{\text{C: reflexive and symmetric but not transitive.}}$

8. Comparison with stored correct answer

Stored correct answer: $\text{C}$.

This matches our derived answer.