Finding the equation of line given the two intersecting lines of planes I have an idea that in order to get the equation given only the lines, it requires to do cross product and solving for the coordinates that satisfies the given two lines.
The problem is that only intersecting lines of planes are given. I have no problem for the cross product but I am confused for finding the coordinates since the answer would be infinite solution. What could be another way to find the points?
The original problem goes this way.

Determine an equation for each of the lines described below.

line of intersection of the planes 2x - y + z = -1 and x + 4y - z = 2.
Crossing the normal vectors to the plane, you’ll get the direction vector of the line, i.e. $$(2,-1,1) \times (1,4,-1) =(-3,3,9) $$
Now, you need any one point that lies on the line. This will be any point that lies on both the planes. One variable is arbitrary. Let’s take $z=0$. Then you get $$2x-y=-1 \\ x+4y =2 \\ \implies x =-\frac 29, y= \frac 59$$ and thus, the equation of the line is $$\vec r = \left( -\frac 29, \frac 59, 0 \right) +\lambda(-3,3,9)$$
Crossing the normal vectors to the plane, you’ll get the direction vector of the line, i.e. $$(2,-1,1) \times (1,4,-1) =(-3,3,9) $$
Now, you need any one point that lies on the line. This will be any point that lies on both the planes. One variable is arbitrary. Let’s take $z=0$. Then you get $$2x-y=-1 \\ x+4y =2 \\ \implies x =-\frac 29, y= \frac 59$$ and thus, the equation of the line is $$\vec r = \left( -\frac 29, \frac 59, 0 \right) +\lambda(-3,3,9)$$
Crossing the normal vectors to the plane, you’ll get the direction vector of the line, i.e. $$(2,-1,1) \times (1,4,-1) =(-3,3,9) $$
Now, you need any one point that lies on the line. This will be any point that lies on both the planes. One variable is arbitrary. Let’s take $z=0$. Then you get $$2x-y=-1 \\ x+4y =2 \\ \implies x =-\frac 29, y= \frac 59$$ and thus, the equation of the line is $$\vec r = \left( -\frac 29, \frac 59, 0 \right) +\lambda(-3,3,9)$$