$a R a \Rightarrow 5 a$ is multiple it 5

So reflexive

$a R b \Rightarrow 2 a +3 b =5 \alpha$,

Now b R a

$2 b+3 a=2 b+\left(\frac{5 \alpha-3 b}{2}\right) \cdot 3$

$=\frac{15}{2} \alpha-\frac{5}{2} b=\frac{5}{2}(3 \alpha-b)$

$=\frac{5}{2}(2 a+2 b-2 \alpha)$

$=5(a+b-\alpha)$

Hence symmetric

$\text { a R b } \quad \Rightarrow 2 a+3 b=5 \alpha \text {. }$

$\text { b R c } \quad \Rightarrow 2 b+3 c=5 \beta$

$\text { Now } \quad 2 a+5 b+3 c=5(\alpha+\beta)$

$\Rightarrow 2 a +5 b +3 c =5(\alpha+\beta)$

$\Rightarrow 2 a +3 c =5(\alpha+\beta- b )$

$\Rightarrow a R c$

Hence relation is equivalence relation.