$\begin{aligned} D & =\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 5 \\ 2 & 3 & \lambda\end{array}\right|=\lambda-6 \\ D_1 & =\left|\begin{array}{ccc}6 & 1 & 1 \\ 10 & 2 & 5 \\ \mu & 3 & \lambda\end{array}\right|=2 \lambda+3 \mu-60\end{aligned}$$$ \begin{aligned} & D_2=\left|\begin{array}{ccc} 1 & 6 & 1 \\ 1 & 10 & 5 \\ 2 & \mu & \lambda \end{array}\right|=4(\lambda-\mu+10) \\ & D_3=\left|\begin{array}{ccc} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 2 & 3 & \mu \end{array}\right|=\mu-16 \end{aligned} $$ For infinite many solution $$ \begin{aligned} & D=D_1=D_2=D_3=0 \\ & \Rightarrow \lambda=6 \& \mu=16 \\ & \Rightarrow \lambda+\mu=22 \end{aligned} $$