We interpret the system as:

For the system to have infinitely many solutions, the third equation must be a linear combination of the first two, and the coefficient matrix must be dependent in a consistent way.

1. Write equations in row form

$R_1=(1,2,3\mid 3),\quad R_2=(4,3,-4\mid 4),\quad R_3=(8,4,-\lambda\mid 9+\mu)$

For infinitely many solutions, we need

$R_3=aR_1+bR_2$

for some constants $a,b$.

2. Compare coefficients of $x$ and $y$

From the $x$-coefficients:

$a+4b=8 \quad ...(1)$

From the $y$-coefficients:

$2a+3b=4 \quad ...(2)$

Solve these:

From (1), $a=8-4b$

Substitute into (2):

$2(8-4b)+3b=4$ $16-8b+3b=4$ $16-5b=4$ $-5b=-12$ $b=\frac{12}{5}$

Then

$a=8-4\cdot \frac{12}{5}=\frac{40-48}{5}=-\frac{8}{5}$

3. Compare coefficient of $z$

Since

$3a-4b=-\lambda$

we get

$3\left(-\frac{8}{5}\right)-4\left(\frac{12}{5}\right)=-\lambda$

$-\frac{24}{5}-\frac{48}{5}=-\lambda$

$-\frac{72}{5}=-\lambda$

Hence,

$\lambda=\frac{72}{5}$

4. Compare constant terms

Also,

$3a+4b=9+\mu$

So,

$3\left(-\frac{8}{5}\right)+4\left(\frac{12}{5}\right)=9+\mu$

$-\frac{24}{5}+\frac{48}{5}=9+\mu$

$\frac{24}{5}=9+\mu$

$\mu=\frac{24}{5}-9=\frac{24}{5}-\frac{45}{5}=-\frac{21}{5}$

5. Final answer

Therefore,

This matches Option D.