analytic geometryconic sections
The solution starts as: Since the plane passes though the vertex, $$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$$ is a generator of the cone. I am not able to see (visualise) why the Normal of the given plane is a generator of the cone. I can see that the plane passes through the vertex.
(Further the solution says since the dr's of the line satisfy cone equation, we plug in $a,b,c$ in C to get $ab+bc+ca=0$ and thus the required result; I am okay with this part)
You can see here for interactive graph
https://www.geogebra.org/3d/ag7rjf5c
First we will make some required transformations to assure that the formula $(b+c)x^2+(c+a)y^2+(a+b)z^2=0\;$ represents a real cone.
A real cone with circular base over the $x \times y$ plane and vertice at the origin should be represented as
$$ K\to m x^2+n y^2+p z^2 = 0 $$
with $m = n$ and $m > 0, n > 0, p < 0$
so we will choose $a,b,c$ such that
$$ b+c=m\\ a+c=n\\ a+b=p $$
in this representation a line $L$
$$ L\to P = \rho\vec v \to \left\{ \begin{array}{rcl} x = \rho \alpha\\ y = \rho \beta\\ z = \rho \gamma \end{array} \right.\;\;\;\; P = (x,y,z) $$
is such that $L \in K$ when
$$ m\alpha^2+n\beta^2+p\gamma^2=0 $$
Now with the help of $\vec v$ and $\vec w = (m\alpha,n\beta,p\gamma)$ Here $\left < \vec v, \vec w\right> = 0$, we can build a plane $\Pi$
$$ \Pi\to P = \lambda \vec w + \mu \left(\vec v \times \vec w\right) $$
as required. Given now the surface
$$ S\to ax^2+by^2+cy^2-1=0\equiv \frac{1}{2} x^2 (-m+n+p)+\frac{1}{2} y^2 (m-n+p)+\frac{1}{2} z^2 (m+n-p)-1 = 0 $$
the intersection $S\cap \Pi$ is obtained
$$ \left(S\circ\Pi\right)(\lambda,\mu) = \frac{1}{2} \left(-4 (m-n-p) (\alpha \lambda m+\beta \gamma \mu (n-p))^2+4 (m+n-p) (\alpha \beta \mu (m-n)+\gamma \lambda p)^2+4 (m-n+p) (\alpha \gamma \mu (p-m)+\beta \lambda n)^2-2\right)= 0 $$
but we can introduce a slight normalization to simplify the representation by considering
$$ m\alpha^2+n\beta^2+p\gamma^2=0\\ \alpha^2+\beta^2+\gamma^2 = 1\\ \beta = \beta_0 = 0\\ m = n\\ $$
and thus we have the intersection curve on $\Pi$ as
$$ 2np^2\lambda^2-2np^2\mu^2-1=0 $$
which is a rectangular hyperbola.
Best Answer
The equation of the cone can be written
$$(x+y+z)^2-(x^2+y^2+z^2)=0$$
Or under a form using a dot product
$$\underbrace{\frac{1}{\sqrt{x^2+y^2+z^2}}\begin{pmatrix}x\\y\\z\end{pmatrix}}_{\text{directing vector of a generatrix}} \ . \ \underbrace{\begin{pmatrix}1\\1\\1\end{pmatrix}}_U=\pm 1 \tag{1}$$
(sign + for the half cone containing $U$, sign - for the other one).
This explains that the cone is circular with axis directed by $U=(1,1,1)$.
Let us denote by $D$ the vector $(a,b,c)$, that we can assume with unit norm WLOG.
(1) is verified in particular for points of type $(x,0,0)$ which means that the $x$ axis is a generator line of the cone. Same thing for the $y$ and the $z$ axes. If we rotate them around $U$, we generate the cone in its totality.
Let $e_x,e_y,e_z$ be the 3 unit vectors of these coordinate axes.
If the 2 intersection lines of the plane with the cone have orthogonal unit norm directing vectors $V_1,V_2$, this pair can be brought by a certain rotation $R$ around $U$ onto the pair $(e_1,e_2)$.
As $D=(a,b,c)$ is orthogonal to $V_1$ and to $V_2$ (this is what plane equation $ax+by+cz=0$ expresses), rotation $R$ sends the line directed by $D=(a,b,c)$ onto the line directed by $e_3$ [just because cross product $D=\pm V_1 \times V_2$ is sent onto $e_3=\pm e_1 \times e_2$]. This means that the line directed by $D$ belongs to the cone, otherwise said
$$\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}$$
is a generator line of the cone.
If it can help you to visualize a little better the situation, see the picture below with the (half) cone as seen from behind as a kind of asian conical hat with orthogonal (red) axes defined by $(e_x,e_y,e_z)$, and $(V_1,V_2,D)$ represented by green axes.