If $\theta$ is the acute angle between the lines represented by the equation $x^2 - 3xy + 2y^2 = 0$, then $\dfrac{3\sin\theta + 2\cos\theta}{3\sin\theta - 2\cos\theta} = $

The given equation of the pair of straight lines is $x^2 - 3xy + 2y^2 = 0$.

Comparing with the standard form $ax^2 + 2hxy + by^2 = 0$, we get $a = 1$, $2h = -3$, and $b = 2$.

The acute angle $\theta$ between the lines is given by $\tan\theta = \left| \dfrac{2\sqrt{h^2 - ab}}{a + b} \right|$.

Substituting the values, we get:

$\tan\theta = \left| \dfrac{2\sqrt{\left(-\dfrac{3}{2}\right)^2 - (1)(2)}}{1 + 2} \right| = \left| \dfrac{2\sqrt{\dfrac{9}{4} - 2}}{3} \right| = \dfrac{2\left(\dfrac{1}{2}\right)}{3} = \dfrac{1}{3}$

We need to find the value of $\dfrac{3\sin\theta + 2\cos\theta}{3\sin\theta - 2\cos\theta}$.

Dividing the numerator and the denominator by $\cos\theta$, we get:

$\dfrac{3\tan\theta + 2}{3\tan\theta - 2}$

Substituting $\tan\theta = \dfrac{1}{3}$:

$\dfrac{3\left(\dfrac{1}{3}\right) + 2}{3\left(\dfrac{1}{3}\right) - 2} = \dfrac{1 + 2}{1 - 2} = \dfrac{3}{-1} = -3$

Answer: $-3$