How to find the point of the line of intersection of two planes I was able to find the direction vector of the line intersecting both planes, but I'm confused on how the textbook obtained the point $(1, 1, -1)$ as the point on the line. The planes are:
$3x + y - z = 3$ and $x - 2y + 4z = -5$
Doing the cross product of the normal vectors made me get: $⟨2, -13, -7⟩$ as the direction vector. The textbook answer for the whole equation is:
$$r =  ⟨1, 1, -1⟩ + t⟨2, -13, -7⟩$$
Please let me know how exactly I can get point $(1, 1, -1)$ as the point on the line.

Notice that every point lying on the line of intersection of two planes must always lie on both the intersecting planes i.e. $3x+y-z=3$ & $x-2y+4z=-5$. We can rewrite them as follows
$$z=3x+y-3\tag 1$$
$$z=\frac{-x+2y}{4}\tag2$$
Solving (1) & (2) by equating, one should get
$$13x+2y=7$$
$$y=\frac{7-13x}{2}\tag3$$
substituting value of $y$ from (3) into (1),
$$z=\frac{1-7x}{2}\tag 4$$
Thus we can take any arbitrary value to $x$ and find the corresponding values of $y$ & $z$   from (3) & (4) respectively.
Therefore there are infinite no. of points lying on the line of intersection.
