When $\Delta=0$ and $\Delta_{1}=\Delta_{2}=\Delta_{3}=0$

then the system of equations has infinite solutions.

-wherein

$a_{1} x+b_{1} y+c_{1} z=d_{1}$

$a_{2} x+b_{2} y+c_{2} z=d_{2}$

$a_{3} x+b_{3} y+c_{3} z=d_{3}$

and

$\Delta=\left|\begin{array}{lll}{a_{1}} & {b_{1}} & {c_{1}} \\ {a_{2}} & {b_{2}} & {c_{2}} \\ {a_{3}} & {b_{3}} & {c_{3}}\end{array}\right|$

$\Delta_{1}, \Delta_{2}, \Delta_{3}$ are obtained by replacing column $1,2,3$ of $\Delta$ by $\left(d_{1}, d_{2}, d_{3}\right)$ column

$\left|\begin{array}{ccc}{1} & {\lambda} & {-1} \\ {\lambda} & {-1} & {-1} \\ {1} & {1} & {-\lambda}\end{array}\right|=0$

$\Rightarrow 1(\lambda+1)-\lambda\left(1-\lambda^{2}\right)-1(1+\lambda)=0$

$\Rightarrow(1+\lambda)\left[\lambda^{2}-\lambda\right]=0$

$\Rightarrow \lambda=-1,0,1$