Find a circular cone that passes through all three…
Find a circular cone that passes through all three coordinate axes.
Explanation of Method
A circular cone containing all three coordinate axes; its axis must make equal angles with them.
Step 1: Axis and half-angle
The three axes pass through a common point, so it is the vertex: $O$. The three axes are generators, so each makes the half-angle $\psi$ with the cone's axis $\vec n=\{p,q,r\}$:
\[ \frac{|p|}{|\vec n|}=\frac{|q|}{|\vec n|}=\frac{|r|}{|\vec n|}=\cos\psi\ \Longrightarrow\ \vec n=\{\pm1,\pm1,\pm1\},\quad \cos\psi=\tfrac1{\sqrt3}. \]
Step 2: Equation
For $\vec n=\{1,1,1\}$, a point $M$ is on the cone if and only if $\cos^2\angle(OM,\vec n)=\tfrac13$: $(x+y+z)^2=x^2+y^2+z^2\iff 2(xy+yz+zx)=0$.
\[xy+yz+zx=0\]
The other three diagonal directions give three more cones: \[xy-yz-zx=0,\qquad -xy+yz-zx=0,\qquad -xy-yz+zx=0,\] with axes $\{1,1,-1\}$, $\{-1,1,1\}$ and $\{1,-1,1\}$.
Each coordinate axis satisfies the equation, because every term contains two different coordinates. $\checkmark$
Result
\[\boxed{\,xy+yz+zx=0\,}\]
Remark
Remark: the source picks the axis $\{1,1,1\}$ "by symmetry" and presents a single answer. The argument above shows why the axis must be a diagonal direction. Up to the choice of diagonal there are 4 such cones, all with half-angle $\arccos\tfrac1{\sqrt3}\approx54.7^\circ$.