To solve for the last three digits of the repetend of $\frac{1}{977}$, we need to determine the digits $A$, $B$, and $C$ from the repeating decimal expansion.

First, we recognize that the decimal expansion of $\frac{1}{977}$ is a repeating decimal with a period of 976 digits. This implies that: $\frac{1}{977} = 0.\overline{d_1 d_2 \ldots d_{976}}$

We want to find the last three digits $d_{974}$, $d_{975}$, and $d_{976}$. Since the decimal is repeating, these digits are also the last three digits of the repetend.

To find these, we use the fact that: $\frac{1}{977} = \frac{A}{10^n - 1} \quad \text{for some integer } A \text{ and } n = 976$

Given that: $0.\overline{d_1 d_2 \ldots d_{976}} = \frac{1}{977}$

we can write the repetend as: $d_1 d_2 \ldots d_{976}$

This periodicity implies: $10^{976} \equiv 1 \pmod{977}$

Now, using modular arithmetic, we need to calculate the multiplicative inverse of $10$ modulo $977$, because the sequence of digits $d_1, d_2, \ldots$ is derived from powers of 10 reduced modulo 977.

We find $10^{975} \pmod{977}$ since we are interested in the last three digits: $10^{975} \times 10 \equiv 1 \pmod{977} \implies 10^{975} \equiv 10^{-1} \pmod{977}$

Next, we need the modular inverse of $10$ modulo $977$. We solve for $x$ in: $10x \equiv 1 \pmod{977}$

Using the Extended Euclidean Algorithm, we find:

977 &= 97 \times 10 + 7, \\ 10 &= 1 \times 7 + 3, \\ 7 &= 2 \times 3 + 1, \\ 3 &= 3 \times 1 + 0. \end{aligned}$$ Working back up: $$\begin{aligned} 1 &= 7 - 2 \times 3, \\ 3 &= 10 - 1 \times 7 \implies 1 = 7 - 2 \times (10 - 1 \times 7) = 3 \times 7 - 2 \times 10, \\ 7 &= 977 - 97 \times 10 \implies 1 = 3 \times (977 - 97 \times 10) - 2 \times 10 \implies 1 = 3 \times 977 - 293 \times 10. \end{aligned}$$ Thus: $$-293 \times 10 \equiv 1 \pmod{977} \implies 10^{-1} \equiv 684 \pmod{977}.$$ Hence: $$10^{975} \equiv 684 \pmod{977}$$ Now we find the last three digits by considering: $$10^{974} \cdot 684 \pmod{977} \equiv 1 \pmod{977}$$ Thus, the last three digits of the repetend of $$\frac{1}{977}$$ are $$A = 6$$, $$B = 8$$, and $$C = 4$$. Therefore, $$A = 6$$, $$B = 8$$, and $$C = 4$$.