QJEEM23Ma369a: JEE Main 2023 Mathematics - Class 12 JEEM23Ma369a Solution: Let be the set of values of , for which the system of equations …

Given that $S$ be the set of values of $\lambda$ for which given system of equations has no solution.

Therefore for the given set of equations

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

$\begin{aligned}&\Rightarrow6 \lambda\left(18 \lambda^2-8\right)+3(6 \lambda-12)+3(4-18 \lambda)=0 \\\\&\Rightarrow18 \lambda^3-14 \lambda-4=0 \\\\&\Rightarrow(\lambda-1)(3 \lambda+1)(3 \lambda+2)=0 \\\\&\Rightarrow \lambda=1,-\frac{1}{3},-\frac{2}{3}\end{aligned}$

Also for each values of $\lambda=1, \frac{-1}{3}, \frac{-2}{3}$ , we have

$\left|\begin{array}{ccc}6 \lambda & -3 & 4 \lambda^2 \\2 & 6 \lambda & 1 \\3 & 2 & \lambda\end{array}\right| \neq 0$

which implies that, for each values of $\lambda$ , the given system of equations has no solution.

$\begin{aligned}& \text { Therefore } S \in\left\{1, \frac{-1}{3}, \frac{-2}{3}\right\} \text { and } \\\\&12 \sum_{\lambda \in S}|\lambda| \\\\& =12\left(|1|+\left|\frac{-1}{3}\right|+\left|\frac{-2}{3}\right|\right) \\\\& =12\left(1+\frac{1}{3}+\frac{2}{3}\right)=12\left(\frac{6}{3}\right)=24\end{aligned}$