Given,

System of equations,

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

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

$3x+2y+3\lambda z=\lambda$

For no solution $△=0$,

So, $\left|\begin{array}{ccc}6\lambda & -3& 3\\ 2& 6\lambda & 4\\ 3& 2& 3\lambda \end{array}\right|=0$

$\Rightarrow 9{\lambda }^3-7\lambda -2=0$

$\Rightarrow (\lambda -1)(3\lambda +1)(3\lambda +2)=0$

Hence, $\lambda \in \left\{1,\frac{-1}{3},\frac{-2}{3}\right\}$

Now putting the value of $\lambda$ in $12\sum _{\lambda \in S}\left|\lambda \right|$ we get,

$12\sum _{\lambda \in S}\left|\lambda \right|=12\times \left(1+\frac{1}{3}+\frac{2}{3}\right)=24$