Intersection points of a line with $xy$ plane Question
A line passing through $(0,0,1)$ and intersecting lines $x+2y+z=1 , -x+y-2z =2$ and $x+y =2 ,x+z=2$. What is the intersecting points of the line with $xy$ plane 
My attempt
I computed the line of intersection of both the pair of plane 
$$L_1= -5i + j + 3k$$
$$L_2 = i - j - k$$
Then tried to get the direction ratios (let it be $ai+bj+ck$) of the line parallel to the line we need to find via the condition that all 3 lines need to be coplanar to intersect. Which gives me
$$a-b+2c=0$$
And I am stuck here 
 A: The plan is:
(1) Find out the parametric equations of the two known lines;
(2) The unknown line intersects the two known lines. This yields two conditions for the parametric equation of the unknown line. Find out these two conditions.
(3) The unknown line passes through the point (0,0,1). This reveals in part the equation of the unknown line. 
(4) The unknown line intersects the xy-plane (or, the plane $z=0$). Using this condition, use the above derived equation of the unknown line to determine the x and y-coordinate of the intersection point. Its z-coordinate is $z=0$. 
In detail,
Line (1) is given as an intersection of two planes,
$$x+2y+z=1$$
$$-x+y-2z=2$$
That is, the first plane has a normal $\vec{n_1}$= (1,2,1), while the second plane has a normal  $\vec{n_2}$= (-1,1,-2).
Then the vector $\vec{v}=\vec{n_1}\times\vec{n_2}$=(-5,-1,3) is parallel to line (1), and its parametric equation is 
$$\vec{x}=\vec{x_0}+\vec{v}t$$
We can see that $\vec{v}$ is not parallel to any of the coordinate planes thus it is safe to assume that line (1) intersects the plane $y=0$. Plug this into the equation of the 2 intersecting planes defining line (1) and find the intersection point to be (4,0,-3). Therefore, the parametric equation of line (1) is:
$$x=4-5t$$
$$y=-t$$
$$z=-3+3t$$
Analogously, process the known equations of the two intersecting planes, defining line (2):
The vector $\vec{v}$ in this case is (1,-1,-1). Line (2) intersects the coordinate plane $y=0$ in point (2,0,0) which gives the parametric equation of line(2) to be:
$$x=2+s$$
$$y=-s$$
$$z=-s$$
Now, assume the unknown line has parametric equation
$$x=x_0+pa$$
$$y=y_0+pb$$
$$z=z_0+pc$$
This line intersects Line (1): at the intersection point, $x$ and $y$ are identical. Therefore:
$$4-5t=x_0+pa$$
$$-t=y_0+pb$$
From here, express $t$, and $p$ through the unknowns $x_0,y_0,a,b$. 
Now, $z$ must be identical for both lines at the intersection point, or else the lines are skew. In the equations for $z$, now plug in $t$ and $p$ to find the first condition:
$$z_0+c\frac{x_0-5y_0-4}{5b-a}=-3-3y_0+3b\frac{x_0-5y_0-4}{a-5b}$$
Analogously, Line (2) and the unknown line intersect and using the same procedure, are found to yield at the intersection point the second condition, namely
$$z_0+c\frac{2-x_0-y_0}{a+b}=y_0-b\frac{x_0+y_0-2}{a+b}$$
Because the unknown line passes through the point (0,0,1), we can write its parametric equation as
$$x=pa$$
$$y=pb$$
$$z=1+pc$$
where $x_0=0, y_0=0,z_0=1$.
Plug these into conditions one and two above. Express $c$. Plug its value into the first condition and solve for $b$:
$$b=\frac{a}{3}.$$
Then, $c=-\frac{a}{3}$. Finally, the parametric eq. of the unknown line is
$$x=pa$$
$$y=\frac{pa}{3}$$
$$z=1-\frac{pa}{3}$$
Use these equations to find out, that this line intersects the coordinate plane $z=0$ at point (3,1,0).
