Find all real values of the real parameter $a$ such that the equation \[ 2x^{2}-6ax+4a^{2}-2a-2+\log_{2}(2x^{2}+2x-6ax+4a^{2})= \] \[ =\log_{2}(x^{2}+2x-3ax+2a^{2}+a+1). \]
has a unique solution.
1. Let's start by simplifying the given equation:
   \[
   2x^{2} - 6ax + 4a^{2} - 2a - 2 + \log_{2}(2x^{2} + 2x - 6ax + 4a^{2}) = \log_{2}(x^{2} + 2x - 3ax + 2a^{2} + a + 1).
   \]

2. Define \( S = 2x^{2} + 2x - 6ax + 4a^{2} \) and \( P = x^{2} + 2x - 3ax + 2a^{2} + a + 1 \). The equation becomes:
   \[
   2x^{2} - 6ax + 4a^{2} - 2a - 2 + \log_{2}(S) = \log_{2}(P).
   \]

3. To simplify further, we need to equate the arguments of the logarithms:
   \[
   2x^{2} - 6ax + 4a^{2} - 2a - 2 + \log_{2}(S) = \log_{2}(P).
   \]

4. Since the logarithm function is strictly increasing, the equation \( \log_{2}(S) = \log_{2}(P) \) implies \( S = P \). Therefore, we have:
   \[
   2x^{2} + 2x - 6ax + 4a^{2} = x^{2} + 2x - 3ax + 2a^{2} + a + 1.
   \]

5. Simplify the equation:
   \[
   2x^{2} + 2x - 6ax + 4a^{2} = x^{2} + 2x - 3ax + 2a^{2} + a + 1.
   \]

6. Subtract \( x^{2} + 2x - 3ax + 2a^{2} + a + 1 \) from both sides:
   \[
   2x^{2} - 6ax + 4a^{2} = x^{2} - 3ax + 2a^{2} + a + 1.
   \]

7. Combine like terms:
   \[
   x^{2} - 3ax + 2a^{2} - a - 1 = 0.
   \]

8. For the quadratic equation \( x^{2} - 3ax + 2a^{2} - a - 1 = 0 \) to have a unique solution, its discriminant must be zero. The discriminant \(\Delta\) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:
   \[
   \Delta = B^2 - 4AC.
   \]

9. Here, \( A = 1 \), \( B = -3a \), and \( C = 2a^{2} - a - 1 \). Therefore:
   \[
   \Delta = (-3a)^2 - 4 \cdot 1 \cdot (2a^{2} - a - 1).
   \]

10. Simplify the discriminant:
    \[
    \Delta = 9a^2 - 4(2a^{2} - a - 1) = 9a^2 - 8a^{2} + 4a + 4 = a^2 + 4a + 4.
    \]

11. Set the discriminant to zero for a unique solution:
    \[
    a^2 + 4a + 4 = 0.
    \]

12. Solve the quadratic equation:
    \[
    (a + 2)^2 = 0.
    \]

13. Therefore:
    \[
    a + 2 = 0 \implies a = -2.
    \]

The final answer is \( \boxed{ a = -2 } \).