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(C) Given the quadratic equation $12x^{2}-20x+3\lambda=0$,the roots $\alpha$ and $\beta$ satisfy $\alpha+\beta = \frac{20}{12} = \frac{5}{3}$ and $\al

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$3$

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(C) Given the quadratic equation $12x^{2}-20x+3\lambda=0$,the roots $\alpha$ and $\beta$ satisfy $\alpha+\beta = \frac{20}{12} = \frac{5}{3}$ and $\alpha\beta = \frac{3\lambda}{12} = \frac{\lambda}{4}$.We are given $\frac{1}{2} \le |\beta-\alpha| \le \frac{3}{2}$.Squaring the inequality,we get $\frac{1}{4} \le (\beta-\alpha)^{2} \le \frac{9}{4}$.Using the identity $(\beta-\alpha)^{2} = (\alpha+\beta)^{2} - 4\alpha\beta$,we have:$\frac{1}{4} \le (\frac{5}{3})^{2} - 4(\frac{\lambda}{4}) \le \frac{9}{4}$.$\frac{1}{4} \le \frac{25}{9} - \lambda \le \frac{9}{4}$.Subtracting $\frac{25}{9}$ from all parts:$\frac{1}{4} - \frac{25}{9} \le -\lambda \le \frac{9}{4} - \frac{25}{9}$.$\frac{9-100}{36} \le -\lambda \le \frac{81-100}{36}$.$-\frac{91}{36} \le -\lambda \le -\frac{19}{36}$.Multiplying by $-1$ reverses the inequality:$\frac{19}{36} \le \lambda \le \frac{91}{36}$.Since $\frac{19}{36} \approx 0.527$ and $\frac{91}{36} \approx 2.527$,the possible integer values for $\lambda$ are $1$ and $2$.The sum of these values is $1+2 = 3$.

List-$I$	List-$II$
$A$. All the roots are negative	$I$. $(b - 3)^2 = 36 + P^2$ for $P \in R$
$B$. Two roots are complex	$II$. $-3 < b < 9$
$C$. Two roots are positive	$III$. $b \in (-\infty, -3) \cup (9, \infty)$
$D$. All roots are real and distinct	$IV$. $b = 9$
	$V$. $b = -3$

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

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