Given $x,y \in \mathbb{R}^3$, $\Vert x \Vert =\Vert y \Vert.$ I want to find all $z \in \mathbb{R}^3$ such that $$\langle x,z \rangle = \langle y,z \rangle.$$ For simplicity, we may use dot product $$\langle x,y\rangle = x_1y_1 + x_2y_2 + x_3 y_3$$
It is obvious when $z = x \times y$ so that $z \perp x$ and $z \perp y.$ Hence, $\langle x,z \rangle = 0 = \langle y,z\rangle.$
But I think there are another $z.$ For example: $$ x = (2,3,0)$$ $$ y = (1, \sqrt{12}, 0)$$ Clearly $\Vert x \Vert = \Vert y \Vert.$ We can choose $z = (0,0,1)$ or in general $z = (0,0, k) k \in \mathbb{R}, k \not = 0$, and hence $\langle x,z\rangle = 0=\langle y,z \rangle.$
But I can find $z = (\sqrt{12}-3, 1, 1)$ or in general $$z = (b(\sqrt{12}-3), b, 1), b \not = 0, b \in \mathbb{R} $$ so that $$ \left (\begin{matrix} 2\\ 3\\ 0 \end{matrix} \right ) \left (\begin{matrix} \sqrt{12}-3 \\ 1\\ 1 \end{matrix} \right )= \left (\begin{matrix} 1 \\ \sqrt{12} \\ 0 \end{matrix} \right ) \left (\begin{matrix} \sqrt{12}-3 \\ 1\\ 1 \end{matrix} \right )$$
Or in general
$$ \left (\begin{matrix}
2\\
3\\
0
\end{matrix} \right ) \left (\begin{matrix}
b(\sqrt{12}-3) \\
b\\
1
\end{matrix} \right )=
\left (\begin{matrix}
1 \\
\sqrt{12} \\
0
\end{matrix} \right ) \left (\begin{matrix}
b(\sqrt{12}-3) \\
b\\
1
\end{matrix} \right ), b \not =0, b \in \mathbb{R}^3.$$
In my opinion, vector $z$ will depend on $x,y$ and so its formula.
In summary:
Given $x,y \in \mathbb{R}^3$, $\Vert x \Vert =\Vert y \Vert.$ I want to find all $z \in \mathbb{R}^3$ such that $$\langle x,z \rangle = \langle y,z \rangle.$$
For early observation, we may use dot product.
Thanks in advanced.
$$\langle x,z \rangle = \langle y,z \rangle$$ is same as $$\langle x-y,z \rangle =0.$$ This means $z$ is orthognal to $x-y$. If $x=y$ then every $z \in \mathbb R^{3}$ satisfies this. Otherwise, the set of all vectors orthogonal to $x-y$ is the plane through the origin with $x-y$ as normal vector.
So the answer is $\{(a,b,c): an_1+bn_2+cn_3=0\}$ where $(n_1,n_2,n_3)=x-y$.