[THOUGHT] To solve this limit problem, I need to evaluate the integral and then take the limit as $n$ approaches infinity. The integrand resembles the form of an exponential function multiplied by a term that looks like a limit definition of the exponential function. I recall that:

Using this, the term $\left(1 + \frac{205}{69n}x\right)^{-69n}$ can be rewritten as:

Thus, the integrand converges to $e^{-\frac{205x}{69}} \cdot e^{-\frac{6}{7}x} = e^{-\left(\frac{205}{69} + \frac{6}{7}\right)x}$.

Now, I need to compute the integral of $e^{-\left(\frac{205}{69} + \frac{6}{7}\right)x}$ from 0 to $\infty$. The integral of $e^{-k x}$ from 0 to $\infty$ is $\frac{1}{k}$, provided $k > 0$.

First, I will compute the coefficient $k = \frac{205}{69} + \frac{6}{7}$. Then, the limit of the integral will be $\frac{1}{k}$. Finally, I will express $\frac{1}{k}$ in reduced form $\frac{p}{q}$ and compute $p^2 q$.

[/THOUGHT] [PYTHON] from fractions import Fraction

Compute the coefficient k

k = Fraction(205, 69) + Fraction(6, 7)

Simplify k

k_simplified = Fraction(k).limit_denominator()

The integral evaluates to 1/k

p_over_q = Fraction(1, k_simplified)

Ensure p and q are coprime

p = p_over_q.numerator q = p_over_q.denominator

Compute p^2 q

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

[THOUGHT] The Python output confirms that $p^2 q = 431351361$. This is the final answer. [/THOUGHT]

$\boxed{431351361}$