If equation rep $x^2+2y^2-5z^2+2kyz+2zx+4xy=0$ represent pair of plane. then $k$ is The values of $k$ for which the equation 
$x^2+2y^2-5z^2+2kyz+2zx+4xy=0$ 
represents a pair of plane passing Through origin,is
what i try
$x^2+2y^2-5z^2+2kyz+2zx+4xy=(ax+by+cz)(px+qy+rz)$
and camparing coefficients
but it is very tedious work
How do i solve it some short way Help me please

We can assume that $a=p=1$, so we have $$x^2+2y^2-5z^2+2kyz+2zx+4xy=(x+by+cz)(x+qy+rz)$$
This should be true for all $y$, so it is true for $y=0$ also, and we get:
 $$x^2-5z^2+2zx=(x+cz)(x+rz)$$  So $cr=-5$ and $c+r=-2$. Also it is true for all $z$ and specialy for $z=0$: $$x^2+2y^2+4xy=(x+by)(x+qy)$$ so $bq=2$ and $b+q=4$...
