$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$
$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$
$2 x+(3 \lambda+1) y+3(\lambda-1) z=0$
has non-zero solutions, is
$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$
$2 x+(3 \lambda+1) y+(3 \lambda-3) z=0$
$\left|\begin{array}{ccc}\lambda-1 & 3 \lambda+1 & 2 \lambda \\ \lambda-1 & 4 \lambda-2 & \lambda+3 \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$
$R _{1} \rightarrow R _{1}- R _{2}$ and $R _{2} \rightarrow R _{2}- R _{3}$
$\left|\begin{array}{ccc}0 & 3-\lambda & \lambda-3 \\ \lambda-3 & \lambda-3 & -2(\lambda-3) \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$
$(\lambda-3)^{2}\left|\begin{array}{ccc}0 & -1 & 1 \\ 1 & 1 & -2 \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$
$(\lambda-3)^{2}[(3 \lambda+1)+(3 \lambda-1)]=0$
$6 \lambda(\lambda-3)^{2}=0 \Rightarrow \lambda=0,3$
$\operatorname{Sum}=3$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\text{f(x)}=1,\ \text{f(y)}\neq1,\ \text{f(z)}\neq2.$
The value of f-1(1)is: