$$ \begin{aligned} & x+2 y+3 z=5 \\\\ & 2 x+3 y+z=9 \\\\ & 4 x+3 y+\lambda z=\mu \end{aligned} $$

For infinite following $\Delta=\Delta_1=\Delta_2=\Delta_3=0$

$\begin{aligned} & \Delta=\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda\end{array}\right|=0 \Rightarrow \lambda=-13 \\\\ & \Delta_1=\left|\begin{array}{llc}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13\end{array}\right|=0 \Rightarrow \mu=15 \\\\ & \Delta_2=\left|\begin{array}{ccc}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & -13\end{array}\right|=0\end{aligned}$

$$ \Delta_3=\left|\begin{array}{ccc} 1 & 2 & 5 \\ 2 & 3 & 9 \\ 4 & 3 & 15 \end{array}\right|=0 $$

For $\lambda=-13, \mu=15$ system of equation has infinite solution hence $\lambda+2 \mu=17$