Equation of plane that goes for intersection of 2 planes and is perpindicular to another plane Really don't know what to do here, went to a tutor neither did he.
Okay first the problem:
Find the equation of the plane that passes through the line of intersection of the planes
x − z = 2 and y + 3z = 1
and is perpendicular to the plane
x + y − 4z = 3
What I have done, not sure if it's right I only have one chance to get it right so I'd really appreciate any input please
< 1, 0, -1 > X < 0, 1, 3 >
which creates the vector: < 1, -3, 1 >
now I crossed it with the normal vector of the third plane:
< 1 , -3 , 1 > X < 1 , 1 , 4>
which creates the vector: < 11, 5 , 4 >
From there I said that if I took a point on the line of intersection and dot product-ed it with the above vector (which is parallel with the plane) then I'd get the equation of the plane.
So I took the two equations and created: 3x + y - 7 = 0
I made the point (1, 4, 0)
So now,
< 11, 5, 4 > dot 
which makes 11x + 5y + 4z - 31 = 0

Thanks for reading if you have, any advice goes a long way, will reply asap
 A: Here is a really neat way of doing problems involving the line of intersection of two planes.  Consider the equation
$$a(x-z-2)+b(y+3z-1)=0\ ,$$
where $a$ and $b$ are constants.  This is the equation of a plane, because it has the right form, namely
$$({\rm constant})x+({\rm constant})y+({\rm constant})z+({\rm constant})=0\ .$$
Also, if $(x,y,z)$ is on the line of intersection of your first two planes, then $x-z=2$ and $y+3z=1$, so this equation is satisfied.  In other words, this is the equation of a plane which contains the line of intersection, and you need to choose $a$ and $b$ in such a way that your third condition is also satisfied.
The normal to this plane is $(a,b,-a+3b)$, and you want this to be perpendicular to $(1,1,-4)$.  Using the dot product gives $5a-11b=0$, and you can take for example $a=11$, $b=5$.  Substitute back into the equation at the top of this answer to get
$$11x+5y+4z-27=0\ .$$
A: Correct normal, incorrect point
The point should be a point on the line of intersection, so if you take $z=0$ then $x=2,y=1$
