$\text{ If the system of linear equations }\mathrm{x}+\mathrm{y}+\mathrm{z}=6,\mathrm{x}+2\mathrm{y}+3\mathrm{z}=14\text{, and }$$2\mathrm{x}+5\mathrm{y}+\mathrm{\lambda z}=\mathrm{\mu }(\mathrm{\lambda },\mathrm{\mu }\in \mathrm{R})\text{ has a unique solution, then }$

The given system of linear equations has a unique solution if$$\begin{gathered} \left|\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\\ 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\lambda \end{array}\right|\ne 0 \\ \Rightarrow 1\left(2\lambda -15\right)-1\left(\lambda -6\right)+1\left(5-4\right)\ne 0 \\ \Rightarrow 2\lambda -15-\lambda +6+1\ne 0 \\ \therefore \lambda -8\ne 0 \end{gathered}$$
$\text{ i.e if }\mathrm{\lambda }-8\ne 0\text{ or }\mathrm{\lambda }\ne 8$