Find an explicit pair of vectors $(u,v)$ in $V$ that span a hyperbolic plane $W$ inside $V$. Consider the symmetric form $\langle\ ,\ \rangle$ on $V=\mathbb{F}_7^3$ defined by the symmetric matrix $$A= \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} \in M_{3 \times 3}(\mathbb{F}_7).$$
Find an explicit pair of vectors $(u,v)$ in $V$ that span a hyperbolic plane $W$ inside $V$.
The symmetric form $\langle\ ,\ \rangle$ is called nondegenerate if the nullspace $N \left(V,\langle\ ,\ \rangle \right)= \{v \in V: \langle v,w \rangle=0 \text{ for every } w \in V \}= \{0\}$.
We call $(V, \langle\ ,\ \rangle)$ a hyperbolic plane if $\dim V = 2$, $\langle\ ,\ \rangle$ is nondegenerate and there is a nonzero isotropic vector $w \in V$.
I'm not really sure where to go here. According to the definitions, we need to find two vectors that span a two-dimensional plane $W$. By the nondegeneracy condition of the symmetric form on $W$, the only vector $v$ such that $\langle v,w \rangle = 0$ is $v=0$. The hyperbolic plane $W$ must also contain a vector $w$ such that $\langle w,w \rangle=0$ in its $\text{span}$. So surely one of the vectors must be perpendicular to itself. How do I proceed? I don't really know how to use the matrix $A$ to obtain the vectors $v,w$ such that $\text{span}(v,w)=W$.

If $\;\begin{pmatrix}x\\y\\z\end{pmatrix}\;$ is isotropic, then $$(x\;y\;z)A\begin{pmatrix}x\\y\\z\end{pmatrix}=(x\;y\;z)\begin{pmatrix}2x\\y+z\\y+2z\end{pmatrix}=2x^2+y^2+2yz+2z^2=0\iff$$
$$\iff2x^2+(y+z)^2+z^2=0$$
The above has only the trivial solution over $\;\Bbb Q\;$ or $\;\Bbb R\;$, say, but over $\;\Bbb C\;$ for example, we have that $\;\left(0,\,1-i,\,i\right)\;$ is an isotropic vector. Over the prime field $\;\Bbb F_7\;$ we have that both
$$\begin{pmatrix}1\\6\\2\end{pmatrix}\;,\;\;\begin{pmatrix}2\\6\\3\end{pmatrix}$$
are isotropic and linearly independ over $\;\Bbb F_7\;%$ .
