(d) : Reflexive : For $(a,b)R(a,b)$\\ $\Rightarrow ab - ab = 0$ is divisible by 5 .\\ So, $(a,b)R(a,b)\forall a,b \in Z$\\ $\therefore\ R$ is reflexive.\\ Symmetric : For $(a,b)R(c,d)$, if $ad - bc$ is divisible by 5 . Then, $bc - ad$ is also divisible by 5 .\\ So, $(c,d)R(a,b)\forall a,b,c,d \in Z$\\ $\therefore\ R$ is symmetric.\\ Transitive : For $(a,b)R(c,d) \Rightarrow ad - bc$ is divisible by 5 and $(c,d)R(e,f) \Rightarrow cf - de$ is divisible by 5\\ Let $ad - bc = 5k_{1}$ and $cf - de = 5k_{2}$, where $k_{1}$ and $k_{2}$ are integers. $\therefore\ adf - bcf = 5k_{1}f\#(i)\$ and $cfb - deb = 5k_{2}b$\\ Solving (i) and (ii), we get $adf - bcf + cfb - deb = 5k_{1}f + 5k_{2}b\#(ii)\$ $\Rightarrow adf - deb = 5\left( k_{1}f + k_{2}b \right) \Rightarrow d(af - be) = 5\left( k_{1}f + k_{2}b \right)$ $af$ - be is not divisible by 5 for $\forall a,b,c,d,e,f \in Z$\\ So, $R$ is not transitive.\\ $\therefore\ R$ is Reflexive and symmetric but not transitive.