Let's call $M$ one matrix associated to $F$.
We know that $F(v) = 0$ for any $v \in \ker(F)$.
This means that each row of $M$ is orthogonal to any vector $v \in \ker(F)$.
A basis of $\ker(F)$ is:
$$k_1 = (2,1,0), k_2 = (-1,0,1).$$
It's easy to see that the vector $h = (1, -2, 1)$ is orthogonal to each vector of the basis of the kernel of $F$. Then any vector proportional to $h$ is orthogonal to any vector $v \in \ker(F)$. Then we can write the matrix $M$ as follows:
$$M = \pmatrix{a & -2a & a \\b & -2b & b \\c & -2c & c \\d & -2d & d},$$
for some real numbers $a, b, c$ and $d$.
Moreover, we know that $F((0,1,1)) = (-1,-3,0,1)$. Then:
$F((0,1,1)) = (-a, -b, -c, -d) = (-1,-3,0,1),$ or equivalently
$$\begin{cases}a = 1 \\b = 3 \\c = 0 \\d = -1\end{cases}$$
yielding to the matrix
$$M = \pmatrix{1 & -2 & 1 \\3 & -6 & 3 \\0 & 0 & 0 \\-1 & 2 & -1}.$$
The answer is the matrix $c$.
