Quadrilateral $ABCD$ is cyclic with $AB = CD = 6$. Given that $AC = BD = 8$ and $AD+3 = BC$, the area of $ABCD$ can be written in the form $\frac{p\sqrt{q}}{r}$, where $p, q$, and $ r$ are positive integers such that $p$ and $ r$ are relatively prime and that $q$ is square-free. Compute $p + q + r$.
1. **Identify the given information and set up the problem:**
   - Quadrilateral \(ABCD\) is cyclic.
   - \(AB = CD = 6\)
   - \(AC = BD = 8\)
   - \(AD + 3 = BC\)

2. **Express \(BC\) in terms of \(AD\):**
   Let \(AD = x\). Then, \(BC = x + 3\).

3. **Apply Ptolemy's Theorem:**
   Ptolemy's Theorem states that for a cyclic quadrilateral \(ABCD\), the sum of the products of its two pairs of opposite sides is equal to the product of its diagonals:
   \[
   AB \cdot CD + AD \cdot BC = AC \cdot BD
   \]
   Substituting the given values:
   \[
   6 \cdot 6 + x \cdot (x + 3) = 8 \cdot 8
   \]
   Simplify the equation:
   \[
   36 + x^2 + 3x = 64
   \]
   \[
   x^2 + 3x + 36 = 64
   \]
   \[
   x^2 + 3x - 28 = 0
   \]

4. **Solve the quadratic equation:**
   \[
   x^2 + 3x - 28 = 0
   \]
   Factorize the quadratic equation:
   \[
   (x + 7)(x - 4) = 0
   \]
   Therefore, the solutions are:
   \[
   x = -7 \quad \text{or} \quad x = 4
   \]
   Since \(x\) represents a length, we discard the negative solution:
   \[
   x = 4
   \]
   Thus, \(AD = 4\) and \(BC = 4 + 3 = 7\).

5. **Calculate the area of the cyclic quadrilateral:**
   We use Brahmagupta's formula for the area of a cyclic quadrilateral:
   \[
   \text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}
   \]
   where \(s\) is the semiperimeter:
   \[
   s = \frac{AB + BC + CD + DA}{2} = \frac{6 + 7 + 6 + 4}{2} = 11.5
   \]
   Substituting the side lengths:
   \[
   \text{Area} = \sqrt{(11.5 - 6)(11.5 - 7)(11.5 - 6)(11.5 - 4)}
   \]
   \[
   = \sqrt{5.5 \cdot 4.5 \cdot 5.5 \cdot 7.5}
   \]
   Simplify the expression inside the square root:
   \[
   = \sqrt{(5.5 \cdot 5.5) \cdot (4.5 \cdot 7.5)}
   \]
   \[
   = \sqrt{30.25 \cdot 33.75}
   \]
   \[
   = \sqrt{1021.875}
   \]
   \[
   = \frac{33\sqrt{15}}{2}
   \]

6. **Express the area in the form \(\frac{p\sqrt{q}}{r}\):**
   \[
   \text{Area} = \frac{33\sqrt{15}}{2}
   \]
   Here, \(p = 33\), \(q = 15\), and \(r = 2\).

7. **Compute \(p + q + r\):**
   \[
   p + q + r = 33 + 15 + 2 = 50
   \]

The final answer is \(\boxed{50}\).