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.
 A: As said in the comments, the given point doesn't work. As for how to get a point on the line, you simply find one solution to the (underdetermined) set of equations given by the two planes. Which is to say the two equations in three unknowns below:
$$
\cases{3x+y-z=3\\x-2y+4z=-5}
$$
There are plenty of ways to get a solution. Maybe the easiest way is to just set one of the unknowns to some value (say set $x=1$) and try to solve the set which now has two unknowns. If that doesn't work, then pick a different unknown instead.
In this case, setting $x=1$ gives
$$
\cases{y-z=0\\-2y+4z=-6}
$$
with the solution $y=z=-3$. So the point $(1,-3,-3)$ lies on the line.
A: To find the intersection, we solve the system :
$$\left\{\begin{array}{lcl} 3 x + y - z & = & 3 \\ x - 2 y + 4 z & = & -5 \end{array}\right.$$
We look for $x$ and $y$ as functions of $z$ :
$$\left\{\begin{array}{lcl} 3 x + y & = & z + 3 \\ x - 2 y & = & -4z - 5 \end{array}\right.$$
then :
$$\left\{\begin{array}{lcl} 7 x & = & 1 - 2 z \\ 7 y & = & 18 + 13 z \end{array}\right.$$
Finally we put $z = 7 t$ to have the equation of the intersection :
$$\left\{\begin{array}{lcl} x & = & \dfrac{1}{7} - 2 t \\ y & = & \dfrac{18}{7} + 13 t \\[3mm] z & = & 7 t \end{array}\right.$$
A: 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.
