The number of 3-digit numbers, that are divisible by either 2 or 3 but not divisible by 7 is _____.

Total 3-digit numbers: $$N = 999 - 100 + 1 = 900$$

Let $$A$$, $$B$$, $$C$$ be sets of 3-digit numbers divisible by 2, 3, 7 respectively.

$$n(A) = \left\lfloor\frac{999}{2}\right\rfloor - \left\lfloor\frac{99}{2}\right\rfloor = 499 - 49 = 450$$

$$n(B) = \left\lfloor\frac{999}{3}\right\rfloor - \left\lfloor\frac{99}{3}\right\rfloor = 333 - 33 = 300$$

$$n(C) = \left\lfloor\frac{999}{7}\right\rfloor - \left\lfloor\frac{99}{7}\right\rfloor = 142 - 14 = 128$$

For intersections:

$$n(A \cap B) \text{ (divisible by 6)} = \left\lfloor\frac{999}{6}\right\rfloor - \left\lfloor\frac{99}{6}\right\rfloor = 166 - 16 = 150$$

$$n(A \cap C) \text{ (divisible by 14)} = \left\lfloor\frac{999}{14}\right\rfloor - \left\lfloor\frac{99}{14}\right\rfloor = 71 - 7 = 64$$

$$n(B \cap C) \text{ (divisible by 21)} = \left\lfloor\frac{999}{21}\right\rfloor - \left\lfloor\frac{99}{21}\right\rfloor = 47 - 4 = 43$$

$$n(A \cap B \cap C) \text{ (divisible by 42)} = \left\lfloor\frac{999}{42}\right\rfloor - \left\lfloor\frac{99}{42}\right\rfloor = 23 - 2 = 21$$

Required numbers: $$n(A \cup B) - n((A \cup B) \cap C)$$

Using set relations: 

$$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 450 + 300 - 150 = 600$$

$$n((A \cup B) \cap C) = n(A \cap C) + n(B \cap C) - n(A \cap B \cap C) = 64 + 43 - 21 = 86$$

$$\text{Required} = 600 - 86 = 514$$