$ \mathrm{A}=\{2,3,6,8,9,11\} \quad(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d}) $

$ \mathrm{B}=\{1,4,5,10,15\} \quad 3 \mathrm{ad}-7 \mathrm{bc} $

$ \text { Reflexive : }(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b})$

$\Rightarrow 3 \mathrm{ab}-7 \mathrm{ba}=-4 \mathrm{ab}$ always even so it is reflexive.

Symmetric : If $3 \mathrm{ad}-7 \mathrm{bc}=$ Even

Case $-I$ : odd odd

Case $-II$ : even even

$(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \Rightarrow 3 \mathrm{bc}-3 \mathrm{ab}$

Case $-I$ : odd odd

Case $-II$ : even even

so symmetric relation

Transitive :

Set $(3,4) R(6,4)$ Satisfy relation

Set $(6,4) R(3,1)$ Satisfy relation

but $(3,4) R(3,1)$ does not satisfy relation so not transitive.