To determine the equation of the normal line to the curve $y = 3 \tan(2x) + 1$ at the point where $x_0 = \frac{\pi}{2}$, we follow these steps:

1. Find the point on the curve: Substitute $x_0 = \frac{\pi}{2}$ into the equation $y = 3 \tan(2x) + 1$.

   $y_0 = 3 \tan\left(2 \cdot \frac{\pi}{2}\right) + 1 = 3 \tan(\pi) + 1 = 3 \cdot 0 + 1 = 1$

   Thus, the point on the curve is $\left(\frac{\pi}{2}, 1\right)$.

2. Find the derivative of the curve: The derivative $y'$ represents the slope of the tangent line. We need to differentiate $y = 3 \tan(2x) + 1$.

   [ y' = \frac{d}{dx} [3 \tan(2x) + 1] = 3 \cdot \frac{d}{dx} [\tan(2x)] = 3 \cdot 2 \sec^2(2x) = 6 \sec^2(2x) ]

   Evaluate the derivative at $x = \frac{\pi}{2}$:

   $y' \bigg|_{x = \frac{\pi}{2}} = 6 \sec^2 \left(2 \cdot \frac{\pi}{2}\right) = 6 \sec^2 (\pi) = 6 \cdot 1 = 6$

3. Find the slope of the normal line: The slope of the normal line is the negative reciprocal of the slope of the tangent line.

   $m_{\text{normal}} = -\frac{1}{y'} = -\frac{1}{6}$

4. Form the equation of the normal line: Use the point-slope form $y - y_1 = m(x - x_1)$ with $\left(\frac{\pi}{2}, 1\right)$ and $m_{\text{normal}} = -\frac{1}{6}$.

   $y - 1 = -\frac{1}{6} \left(x - \frac{\pi}{2}\right)$

   Simplify to get the equation in standard form:

   $y - 1 = -\frac{1}{6}x + \frac{\pi}{12}$ $6y - 6 = -x + \frac{\pi}{2}$ $x + 6y - \frac{\pi}{2} - 6 = 0$ $x + 6y - \frac{\pi}{2} - 6 = 0 \quad \text{or equivalently} \quad x + 6y - 6 - \frac{\pi}{2} = 0$

Now, comparing this with the provided options, we need to clear the denominators and match the format. The correct option is:

$2x + 12y - 12 - \pi = 0$

So, the correct answer is: $\boxed{2x + 12y - 12 - \pi = 0}$

Would you like more details or have any questions about this solution?

Here are some related questions you might find helpful:

1. How do you find the tangent line to a curve at a given point?
2. What is the difference between the tangent and the normal lines to a curve?
3. How do you differentiate trigonometric functions?
4. Can you explain the geometric interpretation of the derivative?
5. What are the applications of normal and tangent lines in real-world problems?

Tip: When finding the equation of a normal line, remember to always use the negative reciprocal of the tangent's slope.