Find the equation to the cone which passes through the three coordinate axes and the lines $$\frac{x}{1}=\frac{y}{-2}=\frac{z}{3}$$ and $$\frac{x}{3}=\frac{y}{2}=\frac{z}{-1}$$ Above is the question from by exercise book, I understand the problem but I cannot reach to the answer. The general equation of a cone with vertex at origin is given by $$ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0$$ It is given that the three coordinate axes are generating line of this cone. Therefore the direction number of the of these lines will satisfy the equation of the cone. The direction number of x-axis is $(1,0,0)$, from here we will get that $a=0$. Similarly, we will get that $b=0$ and $c=0$. Therefore now the equation of the required cone is reduced to $$2fyz+2gzx+2hxy=0$$ Since the cone passes through two more lines and there direction number is given, so we will get the following two equations: $$-6f+3g-2h=0$$ and $$-2f-3g+6h=0$$ Now the problem is that I have only two equations and three unknows. So how to find the values of $f,g,h$?? Please help. Login To add answer/comment