Solution:

Let's find a closed form expression for each of the repeating decimals. We can write $0.\overline{\underline{a}~\underline{b}~\underline{c}}$ as $$0.\underline{a} \ \underline{b} \ \underline{c} + 0.000 \ \underline{a} \ \underline{b} \ \underline{c} + \cdots.$$

From this, we can see that this an infinite geometric sequence with first term $0.\underline{a} \ \underline{b} \ \underline{c}$ and ratio $\dfrac{1}{1000}.$

Using the formula for the sum of a infinite geometric sequence, we get that this equals $$\dfrac{0.\underline{a} \ \underline{b} \ \underline{c}}{1 - \dfrac{1}{1000}} = \dfrac{\underline{a} \ \underline{b} \ \underline{c}}{999}.$$

Similarly, note that we can write $0.\overline{a}$ as $$0.a + 0.0a + 0.00a + \cdots.$$

As above, this equals $$\dfrac{0.a}{1 - \dfrac{1}{10}} = \dfrac{a}{9}.$$ Therefore, $$0.\overline{b} = \dfrac{b}{9} \text{ and } 0.\overline{c} = \dfrac{c}{9}.$$

Substituting all these values into the condition, we get $$\dfrac{\underline{a} \ \underline{b} \ \underline{c}}{999} = \dfrac{1}{3} \cdot \dfrac{a + b + c}{9}.$$ Multiplying through by $999$ yields $$\underline{a} \ \underline{b} \ \underline{c} = 37(a + b +c).$$

Note that we can express $\underline{a} \ \underline{b} \ \underline{c}$ as $100a + 10b + c.$ Substituting this in, we get $$100a + 10b + c = 37(a + b + c),$$ which simplifies to $$63a = 27b + 36c \Rightarrow 7a = 3b + 4c.$$

All the solutions where $a = b = c$ work. The expression $3b + 4c$ remains constant if we increase $b$ by $4$ and decrease $c$ by $4.$ We could also decrease $b$ by $4$ and increase $c$ by $3.$

Applying this principles to the first $9$ triples yields $$(4, 8, 1), (5, 1, 8), (5, 9, 2), (6, 2, 9)$$ as $4$ more solutions. Therefore, there are a total of $13$ solutions.

Thus, D is the correct solution.