$6 \lambda x-3 y+3 z=4 \lambda^2$
$2 x+6 \lambda y+4 z=1,$
$3 \mathrm{x}+2 \mathrm{y}+3 \lambda \mathrm{z}=\lambda$ का कोई हल नहीं है, का समुच्चय $\mathrm{S}$ है, तो $12 \sum_{\lambda \in \mathrm{S}}|\lambda|$ बराबर है___________|
$2 \lambda\left(9 \lambda^2-4\right)+(3 \lambda-6)+(2-9 \lambda)=0$
$18 \lambda^3-14 \lambda-4=0$
$(\lambda-1)(3 \lambda+1)(3 \lambda+2)=0$
$\Rightarrow \lambda=1,-1 / 3,-2 / 3$
For each $\lambda, \Delta_1=\left|\begin{array}{ccc}6 \lambda & -3 & 4 \lambda^2 \\ 2 & 6 \lambda & 1 \\ 3 & 2 & \lambda\end{array}\right| \neq 0$
Ans. $12\left(1+\frac{1}{3}+\frac{2}{3}\right)=24$
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