let $$\alpha, \beta$$ be the roots of the quadratic equation $$12x^{2}-20x+3\lambda=0, \lambda\in \mathbb{Z}$$. If $$\frac{1}{2}\leq |\beta-\alpha|\leq\frac{3}{2}$$, then the sum of all possible values of $$\lambda$$ is :

We have the equation: $$12x^{2}-20x+3\lambda=0, \lambda\in Z$$

Sum of the roots: $$\alpha+\beta\ =-\dfrac{b}{a}=-\ \dfrac{\left(-20\right)}{12}=\dfrac{5}{3}$$

Product of the roots: $$\alpha\ \beta\ =\dfrac{c}{a}=\dfrac{3λ}{12}=\dfrac{λ}{4}$$

Now, we know that: $$\left(\beta\ -\alpha\ \right)^2=\left(\beta\ +\alpha\ \right)^2-4\alpha\ \beta\ $$

$$\left(\beta\ -\alpha\ \right)^2=\left(\dfrac{5}{3}\right)^2-4\left(\dfrac{λ}{4}\right)$$

$$\left(\beta\ -\alpha\ \right)^2=\dfrac{25}{9}-λ$$

$$\left(\beta\ -\alpha\ \right)^2=\dfrac{25-9λ}{9}$$

Taking the square root of both sides:

$$\left|\beta\ -\alpha\right|=\dfrac{\sqrt{\left(25-9λ\right)}}{3}$$

Now, we are given that: 

$$\dfrac{1}{2}\le\ \left|\beta\ -\alpha\right|\le\ \dfrac{3}{2}$$

Substituting:

$$\dfrac{1}{2}\le\ \dfrac{\sqrt{\left(25-9λ\right)}}{3}\le\dfrac{3}{2}$$

Multiplying by 3:

$$\dfrac{3}{2}\le\ \sqrt{\left(25-9λ\right)}\le\dfrac{9}{2}$$

Or, $$\dfrac{9}{4}\le\ 25-9λ\le\dfrac{81}{4}$$

Subtracting 25 from all three:

$$-\dfrac{91}{4}\le\ -9λ\le-\dfrac{19}{4}$$

Dividing all three by -9:

$$\dfrac{91}{36}\ge\ \ λ\ge\dfrac{19}{36}$$

Or, we can say, $$0.527 \leq \lambda \leq 2.527$$

The only integers that fall within the range [0.527, 2.527] are: 1 and 2. 

Hence, the sum of all possible values of $$\lambda$$ is 3.