Write the decimals as $\frac{ab}{99}$ and $\frac{abc}{999},$ then clear denominators using $99 = 9 \cdot 11$ and $999 = 27 \cdot 37$

Reducing the equation modulo $11$ forces $a = b;$ then a short linear equation pins down the digits

Writing $ab$ and $abc$ for the two- and three-digit numbers, the decimals equal $\frac{ab}{99}$ and $\frac{abc}{999}.$ Since $99 = 9 \cdot 11$ and $999 = 27 \cdot 37,$ the common denominator is $27 \cdot 37 \cdot 11 = 10989,$ and multiplying the equation by it gives $$\begin{aligned} &111 \cdot ab + 11 \cdot abc \\ &= \frac{33}{37} \cdot 10989 \\ &= 9801. \end{aligned}$$

Modulo $11,$ since $9801 = 11 \cdot 891$ and $111 \equiv 1,$ this forces $ab$ to be divisible by $11,$ so $a = b.$ Then $ab = 11a,$ and dividing the equation by $11$ gives $111a + abc = 891.$ Since $abc = 110a + c,$ this is $221a + c = 891,$ which requires $a = 4$ and $c = 7.$

Thus $a = b = 4,$ $c = 7,$ and the three-digit number $abc$ is $447.$