Find all real values of the parameter $a$ for which the system
\begin{align*}
&1+\left(4x^2-12x+9\right)^2+2^{y+2}=a\\
&\log_3\left(x^2-3x+\frac{117}4\right)+32=a+\log_3(2y+3)
\end{align*}has a unique real solution. Solve the system for these values of $a$.
1. **Analyzing the first equation:**
   \[
   1 + \left(4x^2 - 12x + 9\right)^2 + 2^{y+2} = a
   \]
   Notice that \(4x^2 - 12x + 9 = (2x - 3)^2\). Therefore, the equation becomes:
   \[
   1 + \left((2x - 3)^2\right)^2 + 2^{y+2} = a
   \]
   Simplifying further:
   \[
   1 + (2x - 3)^4 + 2^{y+2} = a
   \]

2. **Analyzing the second equation:**
   \[
   \log_3\left(x^2 - 3x + \frac{117}{4}\right) + 32 = a + \log_3(2y + 3)
   \]
   Let \(z = x^2 - 3x + \frac{117}{4}\). Then the equation becomes:
   \[
   \log_3(z) + 32 = a + \log_3(2y + 3)
   \]
   Using the properties of logarithms, we can rewrite this as:
   \[
   \log_3(z) - \log_3(2y + 3) = a - 32
   \]
   \[
   \log_3\left(\frac{z}{2y + 3}\right) = a - 32
   \]
   Therefore:
   \[
   \frac{z}{2y + 3} = 3^{a - 32}
   \]

3. **Considering the symmetry in the solutions:**
   If \((x, y) = (t, u)\) is a solution, then \((x, y) = (3 - t, u)\) is also a solution. This symmetry implies that the unique solution must occur when \(x = \frac{3}{2}\).

4. **Substituting \(x = \frac{3}{2}\) into the first equation:**
   \[
   1 + \left(4\left(\frac{3}{2}\right)^2 - 12\left(\frac{3}{2}\right) + 9\right)^2 + 2^{y+2} = a
   \]
   Simplifying inside the parentheses:
   \[
   4\left(\frac{9}{4}\right) - 12\left(\frac{3}{2}\right) + 9 = 9 - 18 + 9 = 0
   \]
   Therefore:
   \[
   1 + 0^2 + 2^{y+2} = a
   \]
   \[
   1 + 2^{y+2} = a
   \]

5. **Substituting \(x = \frac{3}{2}\) into the second equation:**
   \[
   \log_3\left(\left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + \frac{117}{4}\right) + 32 = a + \log_3(2y + 3)
   \]
   Simplifying inside the logarithm:
   \[
   \left(\frac{9}{4}\right) - \frac{9}{2} + \frac{117}{4} = \frac{9 - 18 + 117}{4} = \frac{108}{4} = 27
   \]
   Therefore:
   \[
   \log_3(27) + 32 = a + \log_3(2y + 3)
   \]
   Since \(\log_3(27) = 3\):
   \[
   3 + 32 = a + \log_3(2y + 3)
   \]
   \[
   35 = a + \log_3(2y + 3)
   \]

6. **Solving the system of equations:**
   From the first equation:
   \[
   a = 1 + 2^{y+2}
   \]
   From the second equation:
   \[
   35 = a + \log_3(2y + 3)
   \]
   Substituting \(a = 1 + 2^{y+2}\) into the second equation:
   \[
   35 = 1 + 2^{y+2} + \log_3(2y + 3)
   \]
   \[
   34 = 2^{y+2} + \log_3(2y + 3)
   \]

7. **Finding the unique solution for \(y\):**
   We need to solve:
   \[
   2^{y+2} + \log_3(2y + 3) = 34
   \]
   Testing \(y = 3\):
   \[
   2^{3+2} + \log_3(2 \cdot 3 + 3) = 2^5 + \log_3(9) = 32 + 2 = 34
   \]
   This satisfies the equation, so \(y = 3\) is the unique solution.

8. **Finding the corresponding value of \(a\):**
   \[
   a = 1 + 2^{3+2} = 1 + 32 = 33
   \]

The final answer is \( \boxed{ a = 33 } \)