Find equation of the cone through the coordinate axes and lines $\frac{x}{1}=\frac{y}{-2}=\frac{z}{3}$ and $\frac{x}{3}=\frac{y}{2}=\frac{z}{-1}$. 
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.
 A: The equation of a cone is $r^T Q r = 0$, where $r = [x, y, z]^T $ and $Q$ is symmetric, indefinite matrix.  Thus $Q$ is of the form
$Q = \begin{bmatrix} a && b && c \\ b && d && e \\ c && e && f \end{bmatrix} $
We have the following vectors $r$ that satisfy this equation, which are
$ (1, 0, 0), (0, 1, 0), (0, 0, 1), (1,-2, 3), (3, 2, -1) $
Plugging in the first three vectors results in $a = d = f = 0 $
Plugging in the fourth results in $ -4 b + 6 c - 12 e = 0 $
Plugging the last vector results in $ 12 b - 6 c - 4 e = 0 $
The last two equations can be solved for $(b,c,e)$ , they give
$ (b, c, e) = t (2, 10/3, 1)$
Take $t = 3$ , then $(b, c, e) = (6, 10, 3)$
Thus
$Q = \begin{bmatrix} 0 && 6 && 10 \\ 6 && 0 && 3 \\ 10 && 3 && 0 \end{bmatrix} $
And the equation of the cone is
$ 6 xy + 10 xz + 3 yz = 0 $
Check:  Plug in $(1, 0, 0), (0, 1, 0), (0, 0, 1)$ you get equality.  Plug in $(1,-2, 3)$ you get $ -12 + 30 - 18 = 0 $ .  Plug in $(3,2,-1)$ you get $ 36 - 30 - 6 = 0$, so it checks.
A: The coefficients $a, b, c, f, g, h$ are only determined up to proportionality; if you multiply them all by the same amount, you get the same shape. For example, $2x^2 + 3y^2 = 0$ and $4x^2 + 6y^2 = 0$ give you the same shape.
This means you can set one of the coefficients arbitrarily and then solve for the other two. Let's say you set $f = 1$. Then you can solve $3g - 2h = 6$ and $-3g + 6h = 2$. The solution is $(g, h) = \left(\frac{10}{3}, 2\right)$. If you want all your coefficients to be integers, then you could multiply them all by 3 and say $(a, b, c, f, g, h) = (0, 0, 0, 3, 10, 6)$.
A: You don't need $6$ variables to solve. There are only $5$. You have assumed one extra variable. Take the equation to be :
$$x^2+\frac{b}{a}y^2+\frac{c}{a}z^2+\frac{2f}{a}yz+\frac{2g}{a}zx+\frac{2h}{a}xy=0 $$$$\implies x^2+b'y^2+c'z^2+2f'yz+2g'zx+2h'xy=0 ~~(\text{5 variables only})~~$$
Just like if you assume straight line equation to be $ax+by+c=0$, you need to solve for $3$ variables. In fact, you need only $2$, slope and $y ~\text{(or}~ x)$-intercept.
Now you may proceed.
