Prove that the equation of the cone is $yz(\frac{b}{c}+\frac{c}{b})+zx(\frac{c}{a}+\frac{a}{c})+xy(\frac{a}{b}+\frac{b}{a})=0$ The plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ cuts the coordinate axes in $A,B,C$. Prove that the lines passing through the origin and intersecting the circle $ABC$ generate the cone with equation
$$yz(\frac{b}{c}+\frac{c}{b})+zx(\frac{c}{a}+\frac{a}{c})+xy(\frac{a}{b}+\frac{b}{a})=0$$

My Attempt:
The equation of the generator line passing through the origin and direction cosines $l,m,n$ is $\frac{x}{l}=\frac{y}{m}=\frac{z}{n}$
Cone is a surface generated by the lines called generators passes through fixed point called vertx on the fixed curve called guiding curves.
I am not able to find the equation of the guiding curve the circle $ABC$.Hence i could not find out the equation of the cone.
Please help.

I do not know if it is really helpful, but the center $P$ of the circle has (horrible) coordinates $\dfrac{1}{2(a^2b^2+b^2c^2+c^2a^2)}\begin{pmatrix} a^3(b^2+c^2) \\ b^3(c^2+a^2) \\ c^3(a^2+b^2) \end{pmatrix}$.
Indeed, in the plane, we want to find the circle passing through the vertices of the triangle $ABC$. The center $P$ of this circle is the intersection of the perpendicular bisectors of $[AB]$ and $[AC]$. The middle $M$ of $[AB]$ has coordinates $\begin{pmatrix} \frac{a}{2} \\ \frac{b}{2} \\ 0 \end{pmatrix}$. We want the vectors $\overrightarrow{PM}$ and $\overrightarrow{AB}$ to be perpendicular, so we get the equation $ax-by = \dfrac{a^2-b^2}{2}$. Similarly for $[AC]$ we get $ax-cz = \dfrac{a^2-c^2}{2}$. Do not forget the third equation $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$. Thus we have a linear system to solve, and the unique solution is the aforementioned one.
ADDENDUM after edit:
Let $M=\begin{pmatrix} x\\ y\\ z\end{pmatrix}$ and $t=\dfrac{x}{a}+\dfrac{y}{b} + \dfrac{z}{c}$. The line $(OM)$ cuts the plane if and only if $t \neq 0$. If so, the point of intersection is $M'=\dfrac{1}{t} M_0$. Then $M'$ is on the circle if and only if $\|P-M'\|^2 = \|P-A\|^2$. Now we have (with the obvious abuse of notation about norms of vectors and scalar product) \begin{align*} &\|P-M'\|^2 = \|P-A\|^2 \iff P^2-2P\cdot\dfrac{M}{t} + \dfrac{M^2}{t^2} = P^2 - 2aP_1+a^2 \\
&\iff M^2-2tP\cdot M = (a^2-2aP_1)t^2 \\
&\iff x^2+y^2+z^2-\dfrac{\dfrac{x}{a}+\dfrac{y}{b} + \dfrac{z}{c}}{a^2b^2+b^2c^2+c^2a^2}\left(xa^3(b^2+c^2)+yb^3(c^2+a^2)+zc^3(a^2+b^2)\right) = \dfrac{a^2b^2c^2}{a^2b^2+b^2c^2+c^2a^2}\left(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2} + \dfrac{2xy}{ab}+\dfrac{2yz}{bc}+\dfrac{2zx}{ca}\right)\end{align*}
You can check that the terms in $x^2$, $y^2$ and $z^2$ vanish, and that the remaining terms give you the desired equation. For example, when putting all the terms in the RHS, the coefficient of $xy$ is $\dfrac{1}{a^2b^2+b^2c^2+c^2a^2}\left(2abc^2+\dfrac{b^3(c^2+a^2)}{a} + \dfrac{a^3(b^2+c^2)}{b}\right) = \dfrac{1}{a^2b^2+b^2c^2+c^2a^2} \dfrac{2a^2b^2c^2+b^4(c^2+a^2)+a^4(b^2+c^2)}{ab} = \dfrac{a^2+b^2}{ab} = \dfrac{a}{b}+\dfrac{b}{a}$.
I have to admit this is a bit cumbersome...
