点 A 在曲线 $y = x \ln x^2$（$x > 0$）上。曲线在点 A 处的切线平行于直线 $y - 3x = 1$。 曲线 $y = x \ln x^2$ 在点 A 处的法线与直线 $y - 3x = 1$ 交于点 B。

1. Find the gradient of the line $y - 3x = 1$

The line is $y = 3x + 1$, so its gradient is $3$. Since the tangent at A is parallel to this line, the gradient of the tangent at A is $3$.

2. Differentiate the equation of the curve

Rewrite the curve equation as $y = 2x \ln x$ (since $x > 0$).
Using the product rule:
$\frac{dy}{dx} = 2 \ln x + 2x \left(\frac{1}{x}\right) = 2 \ln x + 2$

3. Equate the derivative to the gradient and solve for $x$

$2 \ln x + 2 = 3$
$2 \ln x = 1$
$\ln x = \frac{1}{2}$
$x = e^{\frac{1}{2}} = \sqrt{e}$

4. Find the corresponding $y$-coordinate

Substitute $x = \sqrt{e}$ into the curve equation:
$y = \sqrt{e} \ln(\sqrt{e})^2 = \sqrt{e} \ln e = \sqrt{e}$
The exact coordinates of A are $(\sqrt{e}, \sqrt{e})$.

1. Find the coordinates of A and the gradient of the normal at A (using results from part i)

From part (i), the gradient of the tangent at A is $3$.
Solving $\frac{dy}{dx} = 3$ gives $x = \sqrt{e}$ and $y = \sqrt{e}$.
So, $A$ is $(\sqrt{e}, \sqrt{e})$.
The gradient of the normal at A is $m_n = -\frac{1}{3}$.

2. Form the equation of the normal at A

$y - \sqrt{e} = -\frac{1}{3}(x - \sqrt{e})$

3. Solve simultaneously with the line $y - 3x = 1$ to find the x-coordinate of B

Substitute $y = 3x + 1$ into the equation of the normal:
$3x + 1 - \sqrt{e} = -\frac{1}{3}(x - \sqrt{e})$
Multiply by 3 to clear the fraction:
$9x + 3 - 3\sqrt{e} = -x + \sqrt{e}$

4. Rearrange to express $x$ in the required form

$10x = 4\sqrt{e} - 3$
$x = \frac{1}{10}(4\sqrt{e} - 3)$
Comparing with $\frac{1}{10}(p\sqrt{e} + q)$, we have $p = 4$ and $q = -3$.