Step 1: Find the coordinates of the intersection points \( P \) and \( Q \).
The equation of the line is \( y = \sqrt{3}x - 3 \).
The equation of the parabola is \( y^2 = x + 2 \).
Substitute the expression for \( y \) from the line equation into the parabola equation:
\[
(\sqrt{3}x - 3)^2 = x + 2
\]
\[
3x^2 - 6\sqrt{3}x + 9 = x + 2
\]
\[
3x^2 - (6\sqrt{3} + 1)x + 7 = 0
\]
This is a quadratic equation in \( x \). Let the roots of this equation be \( x_1 \) and \( x_2 \), which are the x-coordinates of the points \( P \) and \( Q \).
Using Vieta's formulas, we have:
\[
x_1 + x_2 = \frac{6\sqrt{3} + 1}{3}
\]
\[
x_1 x_2 = \frac{7}{3}
\]
The corresponding y-coordinates are \( y_1 = \sqrt{3}x_1 - 3 \) and \( y_2 = \sqrt{3}x_2 - 3 \).
So, the points \( P \) and \( Q \) are \( (x_1, \sqrt{3}x_1 - 3) \) and \( (x_2, \sqrt{3}x_2 - 3) \).
Step 2: Calculate the distances \( XP \) and \( XQ \).
The coordinates of point \( X \) are \( (\sqrt{3}, 0) \).
\[
XP^2 = (x_1 - \sqrt{3})^2 + (\sqrt{3}x_1 - 3 - 0)^2 = (x_1 - \sqrt{3})^2 + (\sqrt{3}(x_1 - \sqrt{3}))^2
\]
\[
XP^2 = (x_1 - \sqrt{3})^2 + 3(x_1 - \sqrt{3})^2 = 4(x_1 - \sqrt{3})^2
\]
\[
XP = 2|x_1 - \sqrt{3}|
\]
Similarly,
\[
XQ^2 = (x_2 - \sqrt{3})^2 + (\sqrt{3}x_2 - 3 - 0)^2 = (x_2 - \sqrt{3})^2 + (\sqrt{3}(x_2 - \sqrt{3}))^2
\]
\[
XQ^2 = (x_2 - \sqrt{3})^2 + 3(x_2 - \sqrt{3})^2 = 4(x_2 - \sqrt{3})^2
\]
\[
XQ = 2|x_2 - \sqrt{3}|
\]
Step 3: Calculate the product \( XP \cdot XQ \).
\[
XP \cdot XQ = 4 |(x_1 - \sqrt{3})(x_2 - \sqrt{3})| = 4 |x_1 x_2 - \sqrt{3}(x_1 + x_2) + 3|
\]
Substitute the values of \( x_1 + x_2 \) and \( x_1 x_2 \) from Step 1:
\[
XP \cdot XQ = 4 \left| \frac{7}{3} - \sqrt{3}\left(\frac{6\sqrt{3} + 1}{3}\right) + 3 \right|
\]
\[
XP \cdot XQ = 4 \left| \frac{7}{3} - \frac{18 + \sqrt{3}}{3} + \frac{9}{3} \right|
\]
\[
XP \cdot XQ = 4 \left| \frac{7 - 18 - \sqrt{3} + 9}{3} \right|
\]
\[
XP \cdot XQ = 4 \left| \frac{-2 - \sqrt{3}}{3} \right| = 4 \left( \frac{2 + \sqrt{3}}{3} \right) = \frac{4(2 + \sqrt{3})}{3}
\]