For $R$ to be an equivalence relation, it needs to be reflexive, symmetric and transitive.

Reflexive: $(a, a) \in R$ for all $a \in N$.
So, $3a + \alpha a = 7k$, where $k$ is an integer.
This implies $(3 + \alpha)a = 7k$.
Thus, $3 + \alpha$ must be a multiple of $7$, so $3 + \alpha = 7k_1$.
Therefore, $\alpha = 7k_1 - 3 = 7k_1 + 4$ for some integer $k_1$.
This indicates that when $\alpha$ is divided by $7$, the remainder is $4$.

Symmetric: If $(a, b) \in R$, then $(b, a) \in R$.
Given $3a + (7k - 3)b = 7m$, then $3(a - b) + 7kb = 7m$.
If $3(b - a) + 7ka = 7m$, then $aRb \implies bRa$.
Thus, $R$ will be symmetric for $\alpha = 7k_1 - 3$.

Transitive: If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.
So, $3a + (7k - 3)b = 7k_1$ and $3b + (7k_2 - 3)c = 7k_3$.
Adding $3a + 7kb + (7k_2 - 3)c = 7(k_1 + k_3)$.
Then $3a + (7k_2 - 3)c = 7m$.
Thus, $(a, c) \in R$.
Therefore, $R$ is transitive.

Since $\alpha = 7k - 3 = 7k + 4$, the remainder when $\alpha$ is divided by $7$ is $4$.