Determining the Filter Coefficients of an FIR Filter
The general formula for the frequency response of a causal second-order FIR filter with a pair of complex zeros on the unit circle is

$$H(e^{j\omega})=G(1-e^{j\omega_0}z^{-1})(1-e^{-j\omega_0}z^{-1})\tag{1}$$

where $\omega_0$ is the (normalized) frequency in radians where the zero occurs, and $G$ is a (real-valued) gain constant. If we choose $G=1$ and expand $(1)$ we obtain

$$H(e^{j\omega})=1-(e^{j\omega_0}+e^{-j\omega_0})z^{-1}+z^{-2}=1-2\cos(\omega_0)z^{-1}+z^{-2}\tag{2}$$

In your example we have

$$\omega_0=2\pi\frac{f_0}{f_s}=2\pi\frac{200}{1200}=\frac{\pi}{3}\tag{3}$$

where $f_0$ is the frequency of the zero, and $f_s$ is the sampling frequency. Plugging $(3)$ into $(2)$ with $\cos(\pi/3)=\frac12$ gives

$$H(e^{j\omega})=1-z^{-1}+z^{-2}\tag{4}$$

which corresponds to the impulse response $\{1,-1,1\}$.