Step 1: Condition for logarithm.
For \(\log_b(y)\) to exist, we require:
\[
y>0
\]
So, step by step we must ensure each inside expression is positive.
Step 2: Innermost logarithm.
We need:
\[
\log_5(20x - x^2 - 91)>0
\]
which means:
\[
20x - x^2 - 91>1 \quad \text{since } \log_5(z)>0 \Rightarrow z>1
\]
So,
\[
20x - x^2 - 91>1
\]
\[
20x - x^2>92
\]
\[
x^2 - 20x + 92<0
\]
Step 3: Solve quadratic inequality.
Discriminant:
\[
D = (20)^2 - 4(92) = 400 - 368 = 32
\]
So, roots are:
\[
x = \frac{20 \pm \sqrt{32}}{2} = \frac{20 \pm 4\sqrt{2}}{2} = 10 \pm 2\sqrt{2}
\]
\[
\Rightarrow 10 - 2\sqrt{2}<x<10 + 2\sqrt{2}
\]
Approximately: \(7.17<x<12.83\).
Step 4: Middle logarithm condition.
We need:
\[
\log_3(\log_5(20x - x^2 - 91))>0
\]
\[
\Rightarrow \log_5(20x - x^2 - 91)>1
\]
\[
\Rightarrow 20x - x^2 - 91>5^1 = 5
\]
\[
20x - x^2>96
\]
\[
x^2 - 20x + 96<0
\]
Step 5: Solve second inequality.
Discriminant:
\[
D = 400 - 384 = 16
\]
Roots:
\[
x = \frac{20 \pm 4}{2} = \{8, 12\}
\]
So range:
\[
8<x<12
\]
Step 6: Outermost logarithm condition.
Since \( \log_7(\cdot) \) requires only positive input, and inside is already ensured>0 by Step 4, no further restriction.
Step 7: Final domain.
Combine Step 3 and Step 5:
- Step 3 gave: \(7.17<x<12.83\)
- Step 5 refined it to: \(8<x<12\)
So final domain:
\[
(8, 12)
\]
Final Answer:
\[
\boxed{(8, 12)}
\]