Quadrilateral $ABCD$ is inscribed in circle $\Gamma$. Segments $AC$ and $BD$ intersect at $E$. Circle $\gamma$ passes through $E$ and is tangent to $\Gamma$ at $A$. Suppose the circumcircle of triangle $BCE$ is tangent to $\gamma$ at $E$ and is tangent to line $CD$ at $C$. Suppose that $\Gamma$ has radius $3$ and $\gamma$ has radius $2$. Compute $BD$.
1. **Understanding the Problem:**
   - We have a cyclic quadrilateral \(ABCD\) inscribed in circle \(\Gamma\) with radius 3.
   - Segments \(AC\) and \(BD\) intersect at \(E\).
   - Circle \(\gamma\) passes through \(E\) and is tangent to \(\Gamma\) at \(A\) with radius 2.
   - The circumcircle of \(\triangle BCE\) is tangent to \(\gamma\) at \(E\) and tangent to line \(CD\) at \(C\).

2. **Tangency and Angles:**
   - Since \(\gamma\) is tangent to \(\Gamma\) at \(A\), the tangency implies that the line through \(A\) and the center of \(\gamma\) is also the tangent line to \(\Gamma\) at \(A\).
   - Given the tangency at \(C\), we have \(\angle EBC = \angle ACD\).

3. **Homothety and Equilateral Triangle:**
   - By homothety, the tangency line at \(A\) is parallel to \(CD\). This implies that \(\triangle ACD\) is isosceles with \(AC = AD\).
   - Since \(\angle EBC = \angle ACD\), and \(\angle ACD = \angle CAD\), it follows that \(\angle ABD = \angle CAD\). Thus, \(AD = CD\).

4. **Using Fact 5:**
   - By Fact 5, the angle bisector of \(\angle ABC\) implies that \(BE\) bisects \(\angle ABC\). Therefore, \(AE = 2CE\).

5. **Law of Cosines and Ptolemy's Theorem:**
   - Denote the length of \(BC = y\) and \(AB = 2y\).
   - Using the Law of Cosines in \(\triangle ACD\), we get:
     \[
     AC = AD = CD = \sqrt{7}y
     \]
   - By Ptolemy's Theorem for cyclic quadrilateral \(ABCD\):
     \[
     AC \cdot BD + AD \cdot BC = AB \cdot CD + AD \cdot BC
     \]
     Since \(AC = AD = CD\), we have:
     \[
     \sqrt{7}y \cdot BD + \sqrt{7}y \cdot y = 2y \cdot \sqrt{7}y + \sqrt{7}y \cdot y
     \]
     Simplifying, we get:
     \[
     \sqrt{7}y \cdot BD = 2y \cdot \sqrt{7}y
     \]
     \[
     BD = 3y
     \]

6. **Area and Radius Calculation:**
   - The area of \(\triangle ACD\) is:
     \[
     \text{Area} = \frac{7\sqrt{3}}{4} y^2
     \]
   - Using the formula for the circumradius \(R\) of \(\triangle ACD\):
     \[
     R = \frac{abc}{4A} = \frac{\sqrt{21}y}{3} = 3
     \]
     Solving for \(y\):
     \[
     3y = \frac{9\sqrt{21}}{7}
     \]

The final answer is \( \boxed{ BD = 3y } \).