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(D) For $R$ to be reflexive,$xRx$ must hold for all $x \in N$. $3x + \alpha x = (3 + \alpha)x$ must be a multiple of $7$. This implies $(3 + \alpha)$

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$4$ is the remainder when $\alpha$ is divided by $7$

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(D) For $R$ to be reflexive,$xRx$ must hold for all $x \in N$. $3x + \alpha x = (3 + \alpha)x$ must be a multiple of $7$. This implies $(3 + \alpha)$ must be a multiple of $7$,so $3 + \alpha = 7k \Rightarrow \alpha = 7k - 3 = 7(k-1) + 4$. Thus,when $\alpha$ is divided by $7$,the remainder is $4$. For $R$ to be symmetric,$xRy \Rightarrow yRx$. $3x + \alpha y = 7n_1$ and $3y + \alpha x = 7n_2$. Subtracting these,$(3 - \alpha)(x - y) = 7(n_1 - n_2)$. This condition is satisfied if $3 + \alpha$ is a multiple of $7$. For $R$ to be transitive,$xRy$ and $yRz \Rightarrow xRz$. $3x + \alpha y = 7n_1$ and $3y + \alpha z = 7n_2$. From the first,$\alpha y = 7n_1 - 3x$. Substituting into the second: $3y + \alpha z = 7n_2$. By eliminating $y$,we find that the relation is transitive if $3 + \alpha$ is a multiple of $7$. Therefore,the condition for $R$ to be an equivalence relation is that $4$ is the remainder when $\alpha$ is divided by $7$.

For $\alpha \in N$,consider a relation $R$ on $N$ given by $R = \{(x, y) : 3x + \alpha y \text{ is a multiple of } 7\}$. The relation $R$ is an equivalence relation if and only if:

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