Let $P$ be the quadratic function such that $P(0) = 7$, $P(1) = 10$, and $P(2) = 25$.  If $a$, $b$, and $c$ are integers such that every positive number $x$ less than 1 satisfies
\[
  \sum_{n = 0}^\infty P(n) x^n = \frac{ax^2 + bx + c}{{(1 - x)}^3},
\]
compute the ordered triple $(a, b, c)$.
1. Let \( P(x) = dx^2 + ex + f \). We are given the following conditions:
   \[
   P(0) = 7 \implies f = 7
   \]
   \[
   P(1) = 10 \implies d + e + f = 10 \implies d + e + 7 = 10 \implies d + e = 3
   \]
   \[
   P(2) = 25 \implies 4d + 2e + f = 25 \implies 4d + 2e + 7 = 25 \implies 4d + 2e = 18 \implies 2d + e = 9
   \]

2. We now have the system of linear equations:
   \[
   \begin{cases}
   d + e = 3 \\
   2d + e = 9
   \end{cases}
   \]

3. Subtract the first equation from the second:
   \[
   (2d + e) - (d + e) = 9 - 3 \implies d = 6
   \]
   Substitute \( d = 6 \) into \( d + e = 3 \):
   \[
   6 + e = 3 \implies e = -3
   \]

4. Therefore, the quadratic function is:
   \[
   P(x) = 6x^2 - 3x + 7
   \]

5. We need to find \( a, b, c \) such that:
   \[
   \sum_{n=0}^\infty P(n) x^n = \frac{ax^2 + bx + c}{(1 - x)^3}
   \]
   Substitute \( P(n) = 6n^2 - 3n + 7 \):
   \[
   \sum_{n=0}^\infty (6n^2 - 3n + 7) x^n
   \]

6. Let \( y = \sum_{n=0}^\infty (6n^2 - 3n + 7) x^n \). We need to express \( y \) in terms of \( \frac{1}{(1-x)^3} \).

7. Consider the series:
   \[
   \sum_{n=0}^\infty n x^n = \frac{x}{(1-x)^2}
   \]
   \[
   \sum_{n=0}^\infty n^2 x^n = \frac{x(1+x)}{(1-x)^3}
   \]

8. Using these series, we can write:
   \[
   y = 6 \sum_{n=0}^\infty n^2 x^n - 3 \sum_{n=0}^\infty n x^n + 7 \sum_{n=0}^\infty x^n
   \]
   \[
   y = 6 \cdot \frac{x(1+x)}{(1-x)^3} - 3 \cdot \frac{x}{(1-x)^2} + 7 \cdot \frac{1}{1-x}
   \]

9. Simplify each term:
   \[
   6 \cdot \frac{x(1+x)}{(1-x)^3} = \frac{6x + 6x^2}{(1-x)^3}
   \]
   \[
   -3 \cdot \frac{x}{(1-x)^2} = \frac{-3x}{(1-x)^2} = \frac{-3x(1-x)}{(1-x)^3} = \frac{-3x + 3x^2}{(1-x)^3}
   \]
   \[
   7 \cdot \frac{1}{1-x} = \frac{7(1-x)^2}{(1-x)^3} = \frac{7 - 14x + 7x^2}{(1-x)^3}
   \]

10. Combine all terms:
    \[
    y = \frac{6x + 6x^2 - 3x + 3x^2 + 7 - 14x + 7x^2}{(1-x)^3}
    \]
    \[
    y = \frac{16x^2 - 11x + 7}{(1-x)^3}
    \]

11. Therefore, \( a = 16 \), \( b = -11 \), and \( c = 7 \).

The final answer is \(\boxed{(16, -11, 7)}\).