Show that the system of equations$$\begin{gathered} 3x-y+4z=3 \\ x+2y-3z=-2 \\ 6x+5y+\lambda z=-3 \end{gathered}$$and has atleast one solution for any real number $\lambda$. Find the set of solutions of $\lambda =-5.$

$D=7(\lambda +5);D_1=4(\lambda +5)$

$D_2=-9(\lambda +5);D_3=0$

For a unique solution  $\mathrm{D}\ne 0$

 If $\mathrm{D}\ne 0\Rightarrow \lambda \ne -5$

then $\mathrm{x}=\frac{4}{7};\mathrm{y}=-\frac{9}{7} and \mathrm{z}=0$ is the solution.

If $\lambda =-5\Rightarrow {\mathrm{D}}_1={\mathrm{D}}_2={\mathrm{D}}_3=\mathrm{D}=0$

In this case  $\mathrm{x}=\frac{4-5 \mathrm{K}}{7};\mathrm{y}=\frac{13 \mathrm{K}-9}{7}$ and  $\mathrm{z}=\mathrm{K}$ is the solution where $\mathrm{K}\in \mathrm{R}.$