Find the sum of possible real values of `x` for which the sixth term of `(3^(log_3 sqrt(9^|x-2|))+7^(1/5 log_7 (3^(|x-2|-9))))^7` equals 567.

Answer

To solve the problem, we need to find the sum of possible real values of \( x \) for which the sixth term of the expression

\[ (3^{\log_3 \sqrt{9^{|x-2|}}} + 7^{\frac{1}{5} \log_7 (3^{|x-2|-9})})^7 \]

equals 567.

### Step-by-Step Solution:

1. **Simplify the Expression**:
- Start with \( 3^{\log_3 \sqrt{9^{|x-2|}}} \):
\[ \sqrt{9^{|x-2|}} = 9^{\frac{|x-2|}{2}} = (3^2)^{\frac{|x-2|}{2}} = 3^{|x-2|} \]
Thus, \( 3^{\log_3 \sqrt{9^{|x-2|}}} = 3^{|x-2|} \).

- Now simplify \( 7^{\frac{1}{5} \log_7 (3^{|x-2|-9})} \):
\[ 7^{\frac{1}{5} \log_7 (3^{|x-2|-9})} = (3^{|x-2|-9})^{\frac{1}{5}} = 3^{\frac{|x-2|-9}{5}} \]

- Combine the two parts:
\[ 3^{|x-2|} + 3^{\frac{|x-2|-9}{5}} \]

2. **Set Up the Sixth Term**:
- The expression becomes:
\[ (3^{|x-2|} + 3^{\frac{|x-2|-9}{5}})^7 \]
- We need to find the sixth term of this binomial expansion, which corresponds to \( T_6 \) (where \( r = 5 \)):
\[ T_6 = \binom{7}{5} (3^{|x-2|})^2 \left(3^{\frac{|x-2|-9}{5}}\right)^5 \]

3. **Calculate the Coefficient**:
- The binomial coefficient:
\[ \binom{7}{5} = \frac{7!}{5! \cdot 2!} = 21 \]

4. **Combine the Terms**:
- The term becomes:
\[ 21 \cdot 3^{2|x-2|} \cdot 3^{\frac{5(|x-2|-9)}{5}} = 21 \cdot 3^{2|x-2|} \cdot 3^{|x-2|-9} = 21 \cdot 3^{3|x-2|-9} \]

5. **Set the Equation**:
- Set this equal to 567:
\[ 21 \cdot 3^{3|x-2|-9} = 567 \]
- Divide both sides by 21:
\[ 3^{3|x-2|-9} = \frac{567}{21} = 27 \]
- Since \( 27 = 3^3 \):
\[ 3^{3|x-2|-9} = 3^3 \]

6. **Equate the Exponents**:
- Thus, we have:
\[ 3|x-2| - 9 = 3 \]
- Solving for \( |x-2| \):
\[ 3|x-2| = 12 \implies |x-2| = 4 \]

7. **Find Possible Values of \( x \)**:
- This gives us two cases:
\[ x - 2 = 4 \implies x = 6 \]
\[ x - 2 = -4 \implies x = -2 \]

8. **Sum of Possible Values**:
- The possible values of \( x \) are \( 6 \) and \( -2 \).
- The sum is:
\[ 6 + (-2) = 4 \]

### Final Answer:
The sum of possible real values of \( x \) is \( \boxed{4} \).