For $R$ to be reflexive $\Rightarrow xRx$

$\Rightarrow 3 x +\alpha x =7 x \Rightarrow(3+\alpha) x =7 K$

$\Rightarrow 3+\alpha=7 \lambda \Rightarrow \alpha=7 \lambda-3=7 N +4, K , \lambda, N \in I$

$\therefore$ when $\alpha$ divided by $7$, remainder is $4$.

$R$ to be symmetric $x R y \Rightarrow y R x$

$3 x +\alpha y =7 N _{1}, 3 y +\alpha x =7 N _{2}$

$\Rightarrow(3+\alpha)( x + y )=7\left( N _{1}+ N _{2}\right)=7 N _{3}$

Which holds when $3+\alpha$ is multiple of $7$

$\therefore \alpha=7 N +4 \text { (as did earlier) }$

$R$ to be transitive

$xRy \& yRz \Rightarrow xRz$.

$3 x +\alpha y =7 N _{1} \& 3 y +\alpha z =7 N _{2}$

$3 x +\alpha z =7 N _{3}$

$\therefore 3 x +7 N _{2}-3 y =7 N _{3}$

$\therefore 7 N _{1}-\alpha y +7 N _{2}-3 y =7 N _{3}$

$\therefore 7\left( N _{1}+ N _{2}\right)-(3+\alpha) y =7 N _{3}$

$\therefore(3+\alpha) y =7\,N$

Which is true again when $3+\alpha$ divisible by $7$ when $\alpha$ divided by $7$, remainder is $4$ .