The given eqn can be written as
$$
\phi(x, y, z)=a x^2+b y^2+c z^2+2 u x+2 v y+2 w z+d=0
$$
we convert this eqn into homogeneous of new variable 't' $\phi(x, y, z, t)=a x^2+b y^2+c z^2+2u x t+2 v y t+2 w z t+d t^2=0$
The condition for this eq represent a cone are
$$
\begin{aligned}
& \frac{\partial \phi}{\partial x}=0 \text {, } a x+u=0 \cdots(1) \\
& \frac{\partial \phi}{\partial y}=0 \text {, } b y+v=0\cdots(2) \\
& \frac{\partial \phi}{\partial 2}=0, cz+w=0\cdots (3)\\
& \frac{\partial \phi}{\partial t}=0 \text {, } ux+v y+w z+d=0 \ldots(4)
\end{aligned}
$$
Solving equation (1), (2) and (3)
$\begin{aligned} x & =-\frac{u}{a} \\ y & =\frac{-v}{b} \\ z & =\frac{-w}{c}\end{aligned}$
The given eq n will be represent a cone only when the value of
$x, y, z$ be satisfy en (4)
hence the required condition is
$$
\begin{aligned}
& u \cdot\left(-\frac{u}{a}\right)+v(-v / b)+w(-w / c)+d=0 \\
& \frac{u^2}{a}+\frac{v^2}{b}+\frac{c^2}{c}=d
\end{aligned}
$$
