When does a conic represents a pair of planes? Given a conic equation
$$H:= ax^2+by^2+cz^2+2fyz+2gzx+2hyx=0,$$
What is (are) the conditions on the coefficients that it represents a pair of planes? 
I know that if this is true, then $H$ can be written as 
$$H=(px+qy+rz)(sx+ty+uz)$$
for some constants $p, q, r, s,t, u$. Expanding gives 
$$ps x^2 + qt y^2 + ruz^2 + (pt+qs)xy+ (pu+ rs) xz+ (qu+rt) yz = H,$$
which gives us a system of $6$ equations
\begin{align}
ps &=a \\ 
qt &= b \\ 
ru &= c \\
pt+qs &= 2f \\
pu + rs &= 2g \\
qu+ rt &= 2h. 
\end{align}
The algebra gets a little bit too messy for me and I do not know how to move on. In the solution, it is given that 
$$f^2 \geq bc, \ \ g^2 \geq ac, \ \ h^2 \geq ab $$
and the determinant of some minor of 
\begin{pmatrix} 
a & h & g \\ h & b & f \\ g & f & c 
\end{pmatrix}
are zero. 
 A: We can confirm that 
$$f^2 \geq bc, \ \ g^2 \geq ac, \ \ h^2 \geq ab$$
and the determinant of a certain matrix equal to $0$ are necessary conditions.
Here is how. The quadratic form can be written under a matrix form as follows :
$$\tag{1}\begin{bmatrix}
x&y&z
\end{bmatrix}\underbrace{\begin{bmatrix}
a&h&g\\h&b&f\\g&f&c
\end{bmatrix}}_Q\begin{bmatrix}
x\\y\\z\\
\end{bmatrix}$$
If the quadratic form can be expressed as the product of two linear forms (when we will equal it to zero, we will get the equations of the two planes passing through the origin):
$$2(px+qy+rz)(sx+ty+uz).$$
The equivalent matrix form of this expression is (many thanks to @Will Jagy) :
$$\begin{bmatrix}
x&y&z
\end{bmatrix}\begin{bmatrix}
p\\q\\r\\
\end{bmatrix}\begin{bmatrix}
s&t&u
\end{bmatrix}\begin{bmatrix}
x\\y\\z\\
\end{bmatrix}+\begin{bmatrix}
x&y&z
\end{bmatrix}\begin{bmatrix}
s\\t\\u\\
\end{bmatrix}\begin{bmatrix}
p&q&r
\end{bmatrix}\begin{bmatrix}
x\\y\\z\\
\end{bmatrix}=$$
$$\tag{2}\begin{bmatrix}
x&y&z
\end{bmatrix}\underbrace{\begin{bmatrix}
2ps&(pt+qs)&(pu+rs)\\(pt+qs)&2qt&(qu+rt)\\(pu+rs)&(qu+rt)&2ru\end{bmatrix}}_Q
\begin{bmatrix}
x\\y\\z\\
\end{bmatrix}$$
And yes, comparing (1) and (2), we check that necessarily, for example the "diagonal minor":
$$\begin{vmatrix}
b&f\\f&c
\end{vmatrix}=bc-f^2=4qurt-(qu+rt)^2=-(qu-rt)^2 \leq 0$$
the same for the other expressions...
Remark : As we can write :
$$Q=\begin{bmatrix}
p&s\\q&t\\r&u
\end{bmatrix}
\begin{bmatrix}
s&t&u\\p&q&r
\end{bmatrix},
$$
the rank of $Q$ is at most 2.
Thus, $det(Q)=0$. 
The rank of $Q$ falls to $1$  iff $\begin{bmatrix}
p\\q\\r
\end{bmatrix}$ and $
\begin{bmatrix}
s\\t\\u
\end{bmatrix}
$ are proportional (the case of a double plane).
A: This is a quadratic form, which is associated with the symmetric matrix
$$Q_H=\begin{bmatrix}
a&h&g\\h&b&f\\g&f&c
\end{bmatrix}$$
and the quadric is degenerate if and only if the discriminant of $H$, i.e. $\det Q_H$ is $0$.
