$$R = \{ (x,y):3x + \alpha y$$ is multiple of 7$$\} $$, now R to be an equivalence relation

(1) R should be reflexive : $$(a,a) \in R\,\forall \,a \in N$$

$$\therefore$$ $$3a + a\alpha = 7k$$

$$\therefore$$ $$(3 + \alpha )a = 7k$$

$$\therefore$$ $$3 + \alpha = 7{k_1} \Rightarrow \alpha = 7{k_1} - 3$$

$$ = 7{k_1} + 4$$

(2) R should be symmetric : $$aRb \Leftrightarrow bRa$$

$$aRb:3a + (7k - 3)b = 7\,m$$

$$ \Rightarrow 3(a - b) + 7kb = 7\,m$$

$$ \Rightarrow 3(b - a) + 7ka = 7\,m$$

So, $$aRb \Rightarrow bRa$$

$$\therefore$$ R will be symmetric for $$a = 7{k_1} - 3$$

(3) Transitive : Let $$(a,b) \in R,\,(b,c) \in R$$

$$ \Rightarrow 3a + (7k - 3)b = 7{k_1}$$ and

$$3b + (7{k_2} - 3)c = 7{k_3}$$

Adding $$3a + 7kb + (7{k_2} - 3)\,c = 7({k_1} + {k_3})$$

$$3a + (7{k_2} - 3)\,c = 7\,m$$

$$\therefore$$ $$(a,c) \in R$$

$$\therefore$$ R is transitive

$$\therefore$$ $$\alpha = 7k - 3 = 7k + 4$$