Let $\mathrm{S}$ be the set of values of $\lambda$, for which the system of equations

$6 \lambda x-3 y+3 z=4 \lambda^{2}$,

$2 x+6 \lambda y+4 z=1$,

$3 x+2 y+3 \lambda z=\lambda$ has no solution. Then $12 \sum_\limits{i \in S}|\lambda|$ is equal to ___________.

Given that $S$ be the set of values of $\lambda$ for which given system of equations has no solution. <br/><br/>Therefore for the given set of equations <br/><br/>$$ \Delta=\left|\begin{array}{ccc} 6 \lambda & -3 & 3 \\ 2 & 6 \lambda & 4 \\ 3 & 2 & 3 \lambda \end{array}\right|=0 $$ <br/><br/>$$ \begin{aligned} &\Rightarrow6 \lambda\left(18 \lambda^2-8\right)+3(6 \lambda-12)+3(4-18 \lambda)=0 \\\\ &\Rightarrow18 \lambda^3-14 \lambda-4=0 \\\\ &\Rightarrow(\lambda-1)(3 \lambda+1)(3 \lambda+2)=0 \\\\ &\Rightarrow \lambda=1,-\frac{1}{3},-\frac{2}{3} \end{aligned} $$ <br/><br/>Also for each values of $\lambda=1, \frac{-1}{3}, \frac{-2}{3}$, we have <br/><br/>$$ \left|\begin{array}{ccc} 6 \lambda & -3 & 4 \lambda^2 \\ 2 & 6 \lambda & 1 \\ 3 & 2 & \lambda \end{array}\right| \neq 0 $$ <br/><br/>which implies that, for each values of $\lambda$, the given system of equations has no solution. <br/><br/>$$ \begin{aligned} & \text { Therefore } S \in\left\{1, \frac{-1}{3}, \frac{-2}{3}\right\} \text { and } \\\\ &12 \sum_{\lambda \in S}|\lambda| \\\\ & =12\left(|1|+\left|\frac{-1}{3}\right|+\left|\frac{-2}{3}\right|\right) \\\\ & =12\left(1+\frac{1}{3}+\frac{2}{3}\right)=12\left(\frac{6}{3}\right)=24 \end{aligned} $$