First time I am encountering this type of question so i just tried but not getting whether my logic correct or not.
First let the sinusoidal signal be $X(t)=\cos(2\pi ft)$.
After sampling this signal I will get $X(nT_s)=\cos(2\pi fnT_s)$; now at sampling rate $f_s=18\, \text{kHz}$ it becomes $X(n)=\cos(\frac{n\pi f}{9})$.
Now I am checking optionS $(C)$ (given correct answer is also C) to see if the output of filter will zero or not at $f=12\,\text{kHz}$ and at 12kHz X(n) becomes $X(nT_s)=\cos\left(\frac{4\pi n}{3}\right)$ .
$H(z)=1+z^{-1}+z^{-2} $ can be written as $H(e^{jw})=1+e^{-jw}+e^{-2jw}$, so from eigenvalue concept the value of $H(e^{jw})$ at $w=\frac{4\pi}{3}$ is $H(e^{jw})=1+e^{\frac{-j4\pi }{3}}+e^{\frac{-j8\pi }{3}}=0$, so it's proven that output at 12 kHz will be zero.
Therefore, 12 kHz frequency will not pass.
I am doing by checking options, option B also satisfying this condition.
Is there any alternate method to solve this?