The equation of a curve is $y = 3x \sin 2x + 1$. The point $A(\pi, 1)$ lies on the curve. The normal to the curve at $A$ meets the $x$-axis at $P$ and the $y$-axis at $Q$. Find length of $PQ$.

1. Differentiate the equation of the curve to find the gradient function.

Given $y = 3x \sin 2x + 1$.
Using the product rule and chain rule:
$\frac{dy}{dx} = 3 \sin 2x + 3x (2 \cos 2x)$
$\frac{dy}{dx} = 3 \sin 2x + 6x \cos 2x$

2. Find the gradient of the tangent at point $A(\pi, 1)$.

Substitute $x = \pi$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\Big|_{x=\pi} = 3 \sin(2\pi) + 6\pi \cos(2\pi)$
Since $\sin(2\pi) = 0$ and $\cos(2\pi) = 1$:
Gradient of tangent, $m_t = 3(0) + 6\pi(1) = 6\pi$

3. Find the gradient of the normal at point $A$.

Gradient of normal, $m_n = -\frac{1}{m_t}$
$m_n = -\frac{1}{6\pi}$

4. Find the equation of the normal at $A(\pi, 1)$.

Using $y - y_1 = m_n(x - x_1)$:
$y - 1 = -\frac{1}{6\pi}(x - \pi)$
$y = -\frac{1}{6\pi}x + \frac{1}{6} + 1$
$y = -\frac{1}{6\pi}x + \frac{7}{6}$

5. Find the coordinates of $P$ (x-intercept) and $Q$ (y-intercept).

For $P$ (x-intercept), set $y = 0$:
$0 = -\frac{1}{6\pi}x + \frac{7}{6}$
$\frac{1}{6\pi}x = \frac{7}{6}$
$x = 7\pi$
So, $P(7\pi, 0)$.

For $Q$ (y-intercept), set $x = 0$:
$y = \frac{7}{6}$
So, $Q(0, \frac{7}{6})$.

6. Calculate the length of $PQ$.

Length $PQ = \sqrt{(7\pi - 0)^2 + (0 - \frac{7}{6})^2}$
$PQ = \sqrt{49\pi^2 + \frac{49}{36}}$
$PQ = \sqrt{\frac{1764\pi^2 + 49}{36}}$
$PQ = \frac{7}{6}\sqrt{36\pi^2 + 1}$