Solution:

We can use the formula for the sum of a geometric sequence to rewrite $A_n, B_n,$ and $C_n.$

$$\begin{gather*} A_n = a(11\cdots11) \\ =a(1 + 10 + 10^2 + \cdots + 10^{n - 1}) \\ =a \cdot \dfrac{10^n - 1}{9} \end{gather*}$$

Similarly, we get that $$B_n = b \cdot \dfrac{10^n - 1}{9}$$ and $$C_n = c \cdot \dfrac{10^{2n} - 1}{9}.$$

We can substitute these expressions into our condition to get $$c \cdot \dfrac{10^{2n} - 1}{9} - b \cdot \dfrac{10^n - 1}{9}$$ $$= a^2 \left(\dfrac{10^n - 1}{9}\right)^2.$$

Simplifying yields $$\begin{align*} c(10^n + 1) - b &= a^2 \cdot \dfrac{10^n - 1}{9} \\ 9c(10^n + 1) - 9b &= a^2 \cdot (10^n - 1) \\ (9c - a^2)10^n &= 9b - 9c - a^2. \end{align*}$$

From the last line, we see that $9c - a^2$ and $9b - 9c - a^2$ are constants.

For there to be at least $2$ unique values of $n$ that satisfy the equation, both sides must equal zero.

We can see this by realizing that this equation is linear with respect to $10^n.$ If both sides are non-zero, then there cannot exist $2$ unique solutions to a linear equation.

This tells us that $$9c - a^2 = 0$$ and $$9b - 9c - a^2 = 0.$$

The first equation gives us $$c = \dfrac{a^2}{9}.$$ Plugging this into the second equation gives us $$b = \dfrac{2a^2}{9}.$$

This tells us that $a$ must be divisible by $3.$ This gives us the following triples:$$(3, 2, 1), (6, 8, 4), (9, 18, 9).$$

The last triple is not allowed, so the maximum sum is $$6 + 8 + 4 = 18.$$ Thus, D is the correct answer.