Given,

The system of equations 

$x+2y+3z=3 ......\left(1\right)$

$4x+3y-4z=4 .........\left(2\right)$

$8x+4y-\lambda z=9+\mu .....\left(3\right)$

Now using the operation, equation $\left(1\right)\times 4-\left(2\right)$ we get,

$5y+16z=8 ........\left(4\right)$

Now using the operation, equation $\left(2\right)\times 2-\left(3\right)$ we get,

$2y+\left(\lambda -8\right)z=-1-\mu ........\left(5\right)$

Now using the operation, equation $\left(4\right)\times 2-\left(3\right)\times 5$ we get,

$\left(72-5\lambda \right)z=21+5\mu$

Now given system of equation has infinitely many solutions, so making coefficient of $z$ to be zero and equating $21+5\mu$ with zero we get,

$72-5\lambda =0\Rightarrow \lambda =\frac{72}{5}$ and $21+5\mu =0\Rightarrow \mu =\frac{-21}{5}$

Hence, the ordered pair $\left(\lambda ,\mu \right)$ is equal to $\left(\frac{72}{5},\frac{-21}{5}\right)$.