If the system of equations

$x+2y+3z=3$

$4x+3y-4z=4$

$8x+4y-\lambda z=9+\mu$

has infinitely many solutions, then the ordered pair ($\lambda,\mu$) is equal to :

For infinite many solution, $\Delta=0$ and $\Delta_x=0$ <br/><br/>$$ \begin{aligned} & \Delta=\left|\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 3 & -4 \\ 8 & 4 & -\lambda \end{array}\right|=0 \\\\ & \Rightarrow 1(-3 \lambda+16)-2(-4 \lambda+32)+3(16-24)=0 \\\\ & \Rightarrow 16-3 \lambda+8 \lambda-64-24=0 \Rightarrow 5 \lambda=72 \\\\ & \therefore \lambda=\frac{72}{5} \\\\ & \Delta_x=\left|\begin{array}{ccc} 3 & 2 & 3 \\ 4 & 3 & -4 \\ 9+\mu & 4 & -\lambda \end{array}\right|=0 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow \quad 3(-3 \lambda+16)-2(-4 \lambda+36+4 \mu)+3(16-27-3 \mu)=0 \\\\ & \Rightarrow-9 \lambda+48+8 \lambda-72-8 \mu-33-9 \mu=0 \\\\ & \Rightarrow-\lambda-17 \mu=57 \\\\ & \Rightarrow-17 \mu=57+\lambda \\\\ & \therefore -\mu=\frac{57+\frac{72}{5}}{17}\\\\ & \Rightarrow \mu=\frac{-357}{85}=\frac{-21}{5} \end{aligned} $$ <br/><br/>$\text { Thus, }(\lambda, \mu) \equiv\left(\frac{72}{5}, \frac{-21}{5}\right)$