$ x+2 y+3 z=5 $

$2 x+3 y+z=9 $

$ 4 x+3 y+\lambda z=\mu$

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

$\Delta=\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda\end{array}\right|=0 \Rightarrow \lambda=-13$

$\Delta_1=\left|\begin{array}{ccc}5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13\end{array}\right|=0 \Rightarrow \mu=15$

$\Delta_2=\left|\begin{array}{ccc}1 & 5 & 3 \\ 2 & 9 & 1 \\ 4 & 15 & -13\end{array}\right|=0$

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

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