$2 x_{1}-4 x_{2}+\lambda x_{3}=1$
$x_{1}-6 x_{2}+x_{3}=2$
$\lambda x_{1}-10 x_{2}+4 x_{3}=3$ નો ઉકેલ શક્ય નથી
$=2(3 \lambda+2)(\lambda-3)$
$D_{1}=-2(\lambda-3)$
$D _{2}=-2(\lambda+1)(\lambda-3)$
$D_{3}=-2(\lambda-3)$
When $\overline{\lambda=3}$, then
$D = D _{1}= D _{2}= D _{3}=0$
$\Rightarrow$ Infinite many solution
when $\left|\lambda=-\frac{2}{3}\right|$ then $D _{1}, D _{2}, D _{3}$ none of them
is zero so equations are inconsistant
$\therefore \lambda=-\frac{2}{3}$