\(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)\).

Modulo 3, \(n^3 \equiv n\) (Fermat's little theorem), so the expression \(\equiv a + b + c \pmod 3\). It is divisible by 3 exactly when \(a + b + c \equiv 0 \pmod 3\).

Of 1 to 99, there are 33 numbers in each residue class (0, 1, 2 mod 3). The sum is divisible by 3 when:

• all three are from the same class: \(3 \times {}^{33}C_3\) ways, or
• one is from each class: \(33 \times 33 \times 33 = ({}^{33}C_1)^3\) ways.

P = \(\frac{3 \times {}^{33}C_3 + ({}^{33}C_1)^3}{{}^{99}C_3}\)

Answer: (A)