Condition that a homogeneous second degree equation in three variables representing a cone.

Question

In my geometry course,we are told that a homogeneous equation of $2$nd degree in $x,y,z$ i.e. $ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0$ will represent a cone with vertex at the origin iff the determinant $$ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{vmatrix} $$ is non zero.But if I take the homogeneous equation $x^2+y^2+z^2=0$,then the LHS is positive definite so $x=y=z=0$ is the only solution.So how can it represent a cone,even though the determinant $$ \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{vmatrix} $$ is non zero,the equation is not representing a cone.Please someone help me to clear my doubts about it. Please see the following attachment for reference (from analytical geometry by Shanti Narayan)

Answer 1

This assertion is false because taking $a=1,b=1,c=1,h=0,f=0$ we have

$$ M = \left( \begin{array}{ccc} 1 & 0 & g \\ 0 & 1 & 0 \\ g & 0 & 1 \\ \end{array} \right) $$

with $\det M = 1-g^2$ and we have a cone representation only for $|g| > 1$