Question

The line x=y=z-1 intersects the cone x^2+y^2+z^2+2yz+2zx+2xy=0 at exactly one point

Answer 1

Let $x=y=z-1=t$ hence

$x^2+y^2+z^2+2yz+2zx+2xy=0\Leftrightarrow\\ t^2+t^2+(t+1)^2+2t(t+1)+2(t+1)t+2t^2=0\Leftrightarrow\\ 4t^2+(t^2+2t+1)+4(t^2+t)=0\Leftrightarrow\\ 9t^2+6t+1=0\Leftrightarrow\\ (3t+1)^2=0\Leftrightarrow\\ 3t+1=0\Leftrightarrow\\ t=-\frac13$

hence $x=t=-\frac13,\,y=t=-\frac13,\,z=t+1=\frac23$ therefore $\left(-\frac13,-\frac13,\frac23\right)$ is the only intersection point, QED.