$$\begin{aligned} &\begin{aligned} & 3 x+y+\beta z=3 \\ & 2 x+\alpha y-z=-3 \\ & x+2 y+z=4 \end{aligned}\\ &\text { has infinite solution }\\ &\begin{aligned} & \Rightarrow \Delta=0, \Delta_1=\Delta_2=\Delta_3 \\ & \Delta=0 \Rightarrow\left|\begin{array}{ccc} 3 & 1 & \beta \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{array}\right|=0 \\ & \Delta_2=0 \Rightarrow\left|\begin{array}{ccc} 3 & 3 & \beta \\ 2 & -3 & -1 \\ 1 & 4 & 1 \end{array}\right|=0 \\ & \Rightarrow 3(-3+4)-3(2+1)+\beta(8+3)=0 \\ & \Rightarrow 3-9+11 \beta=0 \\ & \Rightarrow \quad \beta=\frac{6}{11} \\ & \Delta_3=0 \Rightarrow\left|\begin{array}{ccc} 3 & 1 & 3 \\ 2 & \alpha & -3 \\ 1 & 2 & 4 \end{array}\right|=0 \\ & \Rightarrow \quad 3(4 \alpha+6)-1(8+3)+3(4-\alpha)=0 \\ & \quad 12 \alpha+18-11+12-3 \alpha=0 \\ & \quad 9 \alpha=-19 \\ & \quad \alpha=\frac{-19}{9} \end{aligned} \end{aligned}$$

$\therefore \quad 22 \beta-9 \alpha=31$