1. First, understand the problem: We need to find the sum of all prime factors of $N$ given that $\log_3 \log_5 \left(\frac{N}{2}\right) = 6$. 2. Start by isolating the inner logarithm. Set $x = \log_5 \left(\frac{N}{2}\right)$. Then the equation becomes: $$\log_3 x = 6$$ 3. Convert the logarithmic equation to exponential form: $$x = 3^6$$ Since $3^6 = 729$, we have: $$\log_5 \left(\frac{N}{2}\right) = 729$$ 4. Convert this logarithmic equation to exponential form: $$\frac{N}{2} = 5^{729}$$ 5. Multiply both sides by 2 to solve for $N$: $$N = 2 \times 5^{729}$$ 6. Prime factorization of $N$ is straightforward: $$N = 2^1 \times 5^{729}$$ 7. The distinct prime factors of $N$ are $2$ and $5$. 8. Compute the sum of all prime factors: $$2 + 5 = 7$$ **Final answer:** $7$