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To solve for \(a\), we need to ensure the continuity of the function \( f(x) \) at \( x = 3 \).

### Steps to find \(a\):

1. The function is given by:
   \[
   f(x) = \begin{cases}
   5 + \frac{\sin(ax^2 - 9a)}{x - 3} & \text{if } x \neq 3, \\
   8a - 13 & \text{if } x = 3.
   \end{cases}
   \]
   
   Since the problem states that \( f(x) \) is continuous at every point on \( \mathbb{R} \), particularly at \( x = 3 \), we need to ensure that the limit of \( f(x) \) as \( x \to 3 \) equals the value of \( f(x) \) at \( x = 3 \).

2. First, find the limit of \( f(x) \) as \( x \to 3 \) from the expression for \( x \neq 3 \):
   \[
   \lim_{x \to 3} \left( 5 + \frac{\sin(ax^2 - 9a)}{x - 3} \right).
   \]
   To compute this limit, we need to use L'Hôpital's Rule, as the expression inside involves the indeterminate form \( \frac{0}{0} \).

3. The expression \( ax^2 - 9a \) simplifies to \( a(x^2 - 9) = a(x - 3)(x + 3) \). Therefore, the function becomes:
   \[
   \frac{\sin(a(x - 3)(x + 3))}{x - 3}.
   \]
   As \( x \to 3 \), the numerator \( \sin(a(x - 3)(x + 3)) \to 0 \) and the denominator \( x - 3 \to 0 \), so we can apply L'Hôpital's Rule.

4. Differentiate the numerator and the denominator:
   - The derivative of the denominator \( x - 3 \) is 1.
   - The derivative of the numerator \( \sin(a(x - 3)(x + 3)) \) is \( \cos(a(x - 3)(x + 3)) \cdot \text{(derivative of the argument)} \), which gives:
     \[
     \cos(a(x - 3)(x + 3)) \cdot a(2x).
     \]
   As \( x \to 3 \), this simplifies to:
   \[
   \cos(0) \cdot a(6) = 6a.
   \]
   
   Therefore, the limit becomes:
   \[
   \lim_{x \to 3} \left( 5 + \frac{\sin(a(x^2 - 9))}{x - 3} \right) = 5 + 6a.
   \]

5. For continuity, we require that:
   \[
   \lim_{x \to 3} f(x) = f(3).
   \]
   Thus:
   \[
   5 + 6a = 8a - 13.
   \]

6. Solving for \(a\):
   \[
   5 + 6a = 8a - 13,
   \]
   \[
   5 + 13 = 8a - 6a,
   \]
   \[
   18 = 2a,
   \]
   \[
   a = 9.
   \]

### Conclusion:
The value of \( a \) is \( \boxed{9} \).

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Would you like more details on the solution process or have any further questions? Here are some related questions for practice:

1. How do we apply L'Hôpital's Rule in other indeterminate forms?
2. What are the necessary conditions for a function to be continuous?
3. How would you handle a similar problem if the trigonometric function were different (e.g., cosine)?
4. What is the significance of continuity in real-world applications?
5. How can we differentiate trigonometric functions involving composite arguments?

**Tip:** Always check the limit carefully at the point where the function might be discontinuous. Continuity implies the limit must equal the function's value at that point.

What is the value of a in the function given the continuity condition at x = 3?

Learn how to solve for a in a piecewise function with a trigonometric limit using continuity at x = 3. The solution includes applying L'Hopital's Rule and ensuring the limit equals the function's value at the point of interest. Suitable for Grades 10-12 or early university calculus students.

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