Equation of plane that goes for intersection of 2 planes and is perpindicular to another plane

Question

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

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Thanks for reading if you have, any advice goes a long way, will reply asap

Answer 1

Correct normal, incorrect point
The point should be a point on the line of intersection, so if you take $z=0$ then $x=2,y=1$

Answer 2

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 $$11x+5y+4z-27=0\ .$$