"A rectangular box with edges parallel to the axes has one corner at origin and the opposite corner on the plane x+2y+3z=6. What is the maximum possible volume of the box?"
That's the question given and the solution uses a constraint $G(x, y, z)=x+2y+3z-6=0$. Why is the constraint meant to be 0? What is this equation saying?
The problem has solution if the point $(x,y,z)$ is in the first quadrant. If one allows negative $x,y$, for instance, then one gets arbitrarily high volume (look at the image provided by @Eff). On the other hand the volume of the box is $V(x,y,z)=xyz$. The conditions $x\ge0,y\ge0, z\ge0$ bound a compact region of the plane, hence there is indeed a maximum and a minimum volume. Of course the minimum is $0$ for points with one coordinate $0$, hence we assume $xyz\ne0$ for the sequel. Summing up, we must maximize $V$ subject to $f(x,y,z)=x+2y+3z-6=0$ and Lagrange multipliers say that $\nabla V=\lambda\nabla f$. Thus we look for a point in the given plane such that the vector $\nabla V=(yz,xz,xy)$ is proportional to $\nabla f=(1,2,3)$. This means $xz=2yx, xy=3yz$. Since $xyz\ne0$, the two equations simplify to $z=2y, x=3z$, giving the point $(6y,y,2y)$. But this point must be in the plane, hence $6y+2y+3(2y)-6=0$ and $y=3/7$. Thus the maximum volume is $V(18/7,3/7,6/7)=324/343$.