If the system of equations

$$ \begin{aligned} & 2 x+\lambda y+3 z=5 \\ & 3 x+2 y-z=7 \\ & 4 x+5 y+\mu z=9 \end{aligned} $$

has infinitely many solutions, then $\left(\lambda^2+\mu^2\right)$ is equal to :

<p>$$\begin{aligned} &\begin{aligned} & 2 x+\lambda y+3 z=5 \\ & 3 x+2 y-z=7 \\ & 4 x+5 y+\mu z=9 \end{aligned}\\ &\text { For infinite solutions } \Rightarrow \Delta=0=\Delta_1=\Delta_2=\Delta_3 \end{aligned}$$</p> <p>$$\begin{aligned} & \Delta=\left|\begin{array}{lll} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{array}\right|=0 \\ & \Rightarrow-4 \lambda-3 \lambda \mu+4 \mu+31=0 \\ & \Delta_1=\left|\begin{array}{ccc} 5 & \lambda & 3 \\ 7 & 2 & -1 \\ 9 & 5 & \mu \end{array}\right|=0 \Rightarrow-9 \lambda-7 \lambda \mu+10 \mu+76=0 \\ & \Delta_2=\left|\begin{array}{ccc} 2 & 3 & 5 \\ 3 & -1 & 7 \\ 4 & \mu & 9 \end{array}\right|=0 \Rightarrow \mu+5=0 \Rightarrow \mu=-5 \\ & \Delta_3=\left|\begin{array}{lll} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{array}\right|=0 \Rightarrow \lambda+1=0 \Rightarrow \lambda=-1 \end{aligned}$$</p> <p>$$\begin{aligned} &\therefore \text { For infinite solution } \mu=-5 \text { and } \lambda=-1\\ &\begin{aligned} \text { Now } \mu^2+\lambda^2 & =25+1 \\ & =26 \end{aligned} \end{aligned}$$</p>