There are 10 cards, labeled from 1 to 10. Three cards denoted by $ a,\ b,\ c\ (a > b > c)$ are drawn from the cards at the same time.
Find the probability such that $ \int_0^a (x^2 \minus{} 2bx \plus{} 3c)\ dx \equal{} 0$.
1. **Evaluate the integral:**
   \[
   \int_0^a (x^2 - 2bx + 3c) \, dx
   \]
   We can split the integral into three separate integrals:
   \[
   \int_0^a x^2 \, dx - 2b \int_0^a x \, dx + 3c \int_0^a 1 \, dx
   \]
   Evaluating each integral separately:
   \[
   \int_0^a x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^a = \frac{a^3}{3}
   \]
   \[
   \int_0^a x \, dx = \left[ \frac{x^2}{2} \right]_0^a = \frac{a^2}{2}
   \]
   \[
   \int_0^a 1 \, dx = \left[ x \right]_0^a = a
   \]
   Combining these results, we get:
   \[
   \int_0^a (x^2 - 2bx + 3c) \, dx = \frac{a^3}{3} - 2b \cdot \frac{a^2}{2} + 3c \cdot a = \frac{a^3}{3} - ab^2 + 3ac
   \]

2. **Set the integral equal to zero:**
   \[
   \frac{a^3}{3} - ab^2 + 3ac = 0
   \]
   Since \(a > 0\), we can divide the entire equation by \(a\):
   \[
   \frac{a^2}{3} - ab + 3c = 0
   \]
   Rearrange the equation:
   \[
   a^2 - 3ab + 9c = 0
   \]

3. **Solve for possible values of \(a\), \(b\), and \(c\):**
   Since \(a\), \(b\), and \(c\) are integers and \(a > b > c\), we need to find values that satisfy the equation \(a^2 - 3ab + 9c = 0\).

   - For \(a = 9\):
     \[
     81 - 27b + 9c = 0 \implies 9c = 27b - 81 \implies c = 3b - 9
     \]
     Possible values:
     \[
     b = 5, c = 6 \cdot 5 - 9 = 6
     \]
     \[
     b = 4, c = 3 \cdot 4 - 9 = 3
     \]
     \[
     b = 3, c = 3 \cdot 3 - 9 = 0
     \]

   - For \(a = 6\):
     \[
     36 - 18b + 9c = 0 \implies 9c = 18b - 36 \implies c = 2b - 4
     \]
     Possible values:
     \[
     b = 5, c = 2 \cdot 5 - 4 = 6
     \]
     \[
     b = 4, c = 2 \cdot 4 - 4 = 4
     \]
     \[
     b = 3, c = 2 \cdot 3 - 4 = 2
     \]

   - For \(a = 3\):
     \[
     9 - 9b + 9c = 0 \implies 9c = 9b - 9 \implies c = b - 1
     \]
     Possible values:
     \[
     b = 2, c = 2 - 1 = 1
     \]

4. **Count the valid combinations:**
   The valid combinations are:
   \[
   (9, 5, 4), (9, 4, 2), (6, 5, 2), (3, 2, 1)
   \]
   There are 4 valid combinations.

5. **Calculate the total number of combinations:**
   The total number of ways to choose 3 cards out of 10 is:
   \[
   \binom{10}{3} = \frac{10!}{3!(10-3)!} = 120
   \]

6. **Calculate the probability:**
   The probability is:
   \[
   \frac{\text{Number of valid combinations}}{\text{Total number of combinations}} = \frac{4}{120} = \frac{1}{30}
   \]

The final answer is \(\boxed{\frac{1}{30}}\)