If the system of equations

$$ \begin{aligned} & (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\ & \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\ & (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9 \end{aligned}$$

has infinitely many solutions, then $\lambda^2+\lambda$ is equal to

<p>$$\begin{aligned} &\begin{aligned} & (\lambda-1) x+(\lambda-4) y+\lambda z=5 \\ & \lambda x+(\lambda-1) y+(\lambda-4) z=7 \\ & (\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9 \end{aligned}\\ &\text { For infinitely many solutions }\\ &\begin{aligned} & \mathrm{D}=\left|\begin{array}{ccc} \lambda-1 & \lambda-4 & \lambda \\ \lambda & \lambda-1 & \lambda-4 \\ \lambda+1 & \lambda+2 & -(\lambda+2) \end{array}\right|=0 \\ & (\lambda-3)(2 \lambda+1)=0 \\ & \mathrm{D}_{\mathrm{x}}=\left|\begin{array}{ccc} 5 & \lambda-4 & \lambda \\ 7 & \lambda-1 & \lambda-4 \\ 9 & \lambda+2 & -(\lambda+2) \end{array}\right|=0 \\ & 2(3-\lambda)(23-2 \lambda)=0 \\ & \lambda=3 \\ & \therefore \lambda^2+\lambda=9+3=12 \end{aligned} \end{aligned}$$</p>