[THOUGHT] To solve the given limit involving the integral, we recognize that as ( n \to \infty ), the term ( \left(1 + \frac{205}{69n}x\right)^{-69n} ) approaches ( e^{-205x} ). The integrand then simplifies to ( e^{-205x} \cdot e^{-\frac{6}{7}x} = e^{-\frac{1441}{7}x} ). The integral from 0 to ( \infty ) of ( e^{-\frac{1441}{7}x} , dx ) evaluates to ( \frac{7}{1441} ). Since 7 and 1441 are coprime (1441 factors into 11 and 131, both primes), we find ( p = 7 ) and ( q = 1441 ). Thus, ( p^2 q = 7^2 \times 1441 ).

To verify, we compute ( 7^2 \times 1441 ) using Python. [/THOUGHT] [PYTHON]

Calculate p^2 * q where p = 7 and q = 1441

p = 7 q = 1441 result = p**2 * q print(result) [/PYTHON]

[THOUGHT] The limit simplifies to the integral of ( e^{-\frac{1441}{7}x} ) from 0 to ( \infty ), which evaluates to ( \frac{7}{1441} ). Since 7 and 1441 are coprime, ( p = 7 ) and ( q = 1441 ). The Python computation confirms that ( p^2 q = 7^2 \times 1441 = 70609 ). Thus, the final answer is: [ \boxed{70609} ] [/THOUGHT]