Let S be the set of values of λ , for which the system of equations...

Given,

System of equations,

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

$2x + 6\lambday + 4z = 1$

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

For no solution $△ = 0$,

So, \left|$\begin{cases}6\lambda & - 3 & 3 \\ 2 & 6\lambda & 4 \\ 3 & 2 & 3\lambda\end{cases}$\right| = 0$</p><p>$⇒ 9{\lambda}^{3} - 7\lambda - 2 = 0$</p><p>$⇒ ( \lambda - 1 ) ( 3\lambda + 1 ) ( 3\lambda + 2 ) = 0$</p><p>Hence,$\lambda \in \left\{1, \dfrac{- 1}{3}, \dfrac{- 2}{3}\right\}$</p><p>Now putting the value of$\lambda$in$12\mathop{∑}_{\lambda \in S}\left|\lambda\right|$we get,</p><p>$12\mathop{∑}_{\lambda \in S}\left|\lambda\right| = 12 \times \left(1 + \dfrac{1}{3} + \dfrac{2}{3}\right) = 24