Locus formed by point on a line intersecting 3 other lines in 3D I got this particular question from an old test paper...

Consider three lines given by $y-2=z+3=0$; $z-3=x+1=0$; $x-1=y+2=0$. Let $(\alpha,\beta,\gamma)$ be a point lying on a line intersecting the given three lines. Then the locus generated by $(\alpha,\beta,\gamma)$ is
a. $\quad$ $xy+3yz+2xz+6=0$
b. $\quad$ $3xy+yz+2xz+6=0$
c. $\quad$ $2xy+3yz+xy+6=0$
d. $\quad$ none of these

I suppose that the intersecting line is as
$$\frac{x-\alpha}{l}=\frac{y-\beta}{m}=\frac{z-\gamma}{n}$$
where $l,m,n$ are direction ratios.
Now I can't understand how to proceed from here. Please help me out...
 A: Here's something to get you started.
First you need to find the lines that intersect all three of the given lines. To do that, fix a point on the first given line, say $(u,2,-3)$. Now you want to find the point on the second given line, say $(-1,v,3)$, such that the line through those two points intersects the third given line. Find a parameterization of that line. That parameterization will have its own parameter, say $t$, and that parameterization will include your original variable $u$. So now you have a parameterization in two variables of all the points on all those lines.
Then find which of those equations $a,b,c$ in your multiple choice will be satisfied by your parameterization. Then you are done!

Answer to your comment:
For a point $(x,y,z)$ to be on the third line we must have $x=1,y=-2$. Substitute those into your parameterization and you get
$$\frac{1-u}{-1-u}=\frac{-2-2}{v-2}=t$$
Solving for $v$ gives
$$v=2+\frac{4+4u}{1-u}$$
Substituting that into your parameterization and solving for $x,y,z$ gives
$$x=(-u-1)t+u, \quad y=\left(\frac{4+4u}{1-u}\right)t+2, \quad z=6t-3$$
There is the parameterization of your desired surface. Now find which of the three possible equations, if any, are satisfied by that parameterization.
Can you finish from here?
A: The general setting of this question is interesting, let me work out a general formula for the locus first.
First, any line $\ell$ on $\mathbb{R}^3$ can be described by a pair of vectors $(\vec{p}, \vec{t})$ where $\vec{p}$ is a point on $\ell$ and $\vec{t} \ne \vec{0}$ points along its tangent direction:
$$\ell = \{\; \vec{p} + \lambda \vec{t} : t \in \mathbb{R} \;\}$$
To abuse notation, for any line $\ell_?$ labelled by a index '?', we will use 
the notation $\vec{p}_?$ and $\vec{t}_?$ to denote an arbitrary chosen pair of vectors describing that line. Once $\vec{p}_?$ and $\vec{t}_?$ are chosen,
we will use $u_?(\vec{p})$ as a shorthand for the expression $(\vec{p} - \vec{p}_?) \times \vec{t}_?$.
Given any two lines $\ell_1, \ell_2$ not parallel to each other.
 $\vec{t}_1 \times \vec{t}_2$ will be non-zero and perpendicular to both $\vec{t}_1$ and $\vec{t}_2$. If we look at the two lines from a direction perpendicular to this vector, the image of the two lines will become parallel.
Their separation will be proportional to $(\vec{p}_1 - \vec{p}_2)\cdot \vec{t}_1 \times \vec{t}_2$.
A consequence of this is:

Two non-parallel lines $\ell_1$ and $\ell_2$ intersect if and only if
  $$(\vec{p}_1 - \vec{p}_2)\cdot \vec{t}_1 \times \vec{t}_2 = 0
\quad\iff\quad u_1(\vec{p}_2) \cdot \vec{t}_2 = 0
\quad\iff\quad u_2(\vec{p}_1) \cdot \vec{t}_1 = 0.$$

Given any three non-parallel lines $\ell_1, \ell_2, \ell_3$ and a point $\vec{p}$ outside the three lines. If $\vec{p}$ lies on the locus, then one can
find a non-zero vector $\vec{t}$ such that
$$u_1(\vec{p})\cdot \vec{t} = 
  u_2(\vec{p})\cdot \vec{t} = 
  u_3(\vec{p})\cdot \vec{t} = 0
\tag{*1}
$$
Since $\vec{p}$ doesn't lies on these 3 lines, the 3 vectors $u_i(\vec{p})$ are
non-zero. We can find a non-zero $\vec{t}$ to satisfy $(*1)$ when and only when
these 3 vectors are linear dependent to each other which is equivalent to the
vanishing of their triple product. To summarize,

The condition for a point $\vec{p}$ to lie on the locus for three non-parallel
  $\ell_1, \ell_2, \ell_3$ is
  $$u_1(\vec{p}) \cdot ( u_2(\vec{p}) \times u_3(\vec{p}) ) = 0$$

Back to our original problem. 
Let $(x,y,z)$ be the coordinates for a generic point
$\vec{p}$. It is easy to see we can represent the three given lines as
$$\begin{cases}
( \vec{p}_1, \vec{t}_1 ) &= ( (0,2,-3), (1, 0, 0) )\\
( \vec{p}_2, \vec{t}_2 ) &= ( (-1,0,3), (0, 1, 0) )\\
( \vec{p}_3, \vec{t}_3 ) &= ( (1,-2,0), (0, 0, 1) )
\end{cases}
\implies
\begin{cases}
u_1(\vec{p}) = ( 0, z+3, 2-y)\\
u_2(\vec{p}) = ( 3-z,0,x+1)\\
u_3(\vec{p}) = ( y+2,1-x,0)
\end{cases}
$$
The equation of the locus becomes
$$u_1(\vec{p})
\cdot ( u_2(\vec{p})
\times u_3(\vec{p}) ) =
\left|\begin{matrix}
 0  & z+3 & 2-y \\
3-z & 0 & x+1\\
 y+2 &1-x & 0
\end{matrix}\right|
= 6xy + 2yz + 4xz + 12 = 0
$$
So the answer is $(b)$.
