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Solution

For $12x^2-20x+3\lambda=0$, $|\beta-\alpha|=\dfrac{\sqrt{D}}{12}=\dfrac{\sqrt{400-144\lambda}}{12}=\dfrac13\sqrt{25-9\lambda}$. Given $\dfrac12\le \dfrac13\sqrt{25-9\lambda}\le \dfrac32$. So $\dfrac32\le \sqrt{25-9\lambda}\le \dfrac92$. Squaring gives $\dfrac94\le 25-9\lambda\le \dfrac{81}{4}$. Hence $\dfrac{19}{36}\le \lambda\le \dfrac{91}{36}$. Since $\lambda\in\mathbb Z$, possible values are $1,2$. Their sum is $3$.