Let A = \ n ∈ N ∣ n ^ 2 ≤ n + 10, 000\ , B = \ 3k + 1 ∣ k ∈ N\ and...

We have, $A = \left\{n \in N ∣ {n}^{2} \le n + 10, 000\right\}, B = \left\{3k + 1 ∣ k \in N\right\}$ and $C = \left\{2k ∣ k \in N\right\},$

$B ≡ \left\{4, 7, 10, 13, 16, 19, \ldots \ldots\right\}$

$C ≡ \left\{2, 4, 6, 8, 10, 12, 14, 16, 20, \ldots \ldots\right\}$

$B - C ≡ \left\{7, 13, 19, \ldots 97, \ldots\right\}$

Now, ${n}^{2} - n \le 100 \times 100$

$⇒ n\left(n - 1\right) \le 100 \times 100$

$⇒ A = \left\{1, 2, \ldots , 100\right\}$

So, $A \cap \left(B - C\right) = \left\{7, 13, 19, \ldots , 97\right\}$

Let ${S}_{n} = 7 + 13 + 19 + \ldots + 97$

Clearly, the terms are in arithmetic progression. Here, $a = 7$ and $d = 13 - 7 = 6$ and $l = 97$

$⇒ l = a + \left(n - 1\right)d = 97$ where $n$ is the number of terms.

$⇒ 7 + \left(n - 1\right)6 = 97$

$⇒ \left(n - 1\right)6 = 90$

$⇒ \left(n - 1\right) = 15$

$⇒ n = 16$

Hence, sum ${S}_{n} = 7 + 13 + 19 + \ldots + 97$

${S}_{n} = \dfrac{n}{2}\left(a + l\right)$

${S}_{n} = \dfrac{16}{2}\left(7 + 97\right)$

$= 832 .$