Describe all planes perpendicular to a plane, and all lines parallel to two given planes. 


*

*I have found two planes trough the origin that meet the given plane
at right angles.


I found three points in the plane, getting the vectors between those points, and using that as a normal vector for the equation of the plane I am looking to find. 
I don't understand how I am to describe all of these planes though. Would finding a basis of all vectors lying in the plane and using an arbitrary linear combination of these as our normal to the plane we are looking for be the right way? 


*I took $(4, -1, 1)$ and $(1, -2, -3)$ as normal vectors to each plane.
Their cross product is $(5, 13, -7)$ is in the direction of the line
of intersection of the two planes, thus parallel to both.


$(x, y, z) = (x_0 + 5t, y_0 + 13t, z_0 -7t)$ is a parametric equation for all lines parallel to both. Is this correct? And what do they mean by refining this such that each line is listed once?
Any help is greatly appreciated!
 A: The equation $4x-8y+11z = 0$ can be rewritten as $\langle (x,y,z) , (4,-8,11)\rangle = 0$.
In this form, we can see that the plane is the set of points $v = (x,y,z)$ that are perpendicular to $u = (4,-8,11)$.
In other words, the set of $v$ with $v\perp u$; call this plane $P_u$.
Any plane in $\Bbb R^3$ can be described in this manner.
A plane $P_w$ is perpendicular to $P_u$ $\iff$ $w\perp u$ with $w\neq 0$.
This describes all the planes via the inner product equation above, and also tells you how to find two planes that are perpendicular to each other and also to $P_u$.

Your procedure for answering $(b)$ is correct.
Now, notice that, varying $p_0 =(x_0,y_0,z_0)$, the same line can show up multiple times.
Indeed, notice that the set
$$\{(x_0-t,y_0 - 11t, z_0 - 7t)\in\Bbb R^3\,|\, t\in\Bbb R\}$$
coincides with, for instance,
$$\left\{\big((x_0 - 1)-t,(y_0-11) - 11t, (z_0 - 7) - 7t\big)\in\Bbb R^3\,|\, t\in\Bbb R\right\}.$$
The first is obtained with some $p_0$ as the inital point and the second is obtained from the first with $p_0 + (- 1, - 11, -7)$.
Visually, you're picking an anchor point $p_0$, and the parameter $t$ defines how much one moves along the direction vector $(-1,-11,-7)$.
Clearly, any anchor other point on the same line will not change the set of points one can reach in this way.
Do you think you can take it from here?
A: For (a), I agree the form of the desired answer for "all planes" is a little unclear. You might find some clues in earlier exercises.
Your idea about finding a basis of the original plane and taking linear combinations seems like a reasonable approach.
You could even use a single parameter $\theta$ to specify the linear combination like this:
$$ (\cos\theta) u + (\sin\theta) v $$
where $u$ and $v$ are your basis vectors.
(Note: $\theta$ will not measure the angle of the plane to any reference line unless the basis is orthonormal. Personally I don't see why that would be important.)
For (b), I agree with your answer so far (especially after the edit that fixed a sign error).
Note that $(x, y, z) = ( 5 t,  13 t,  -7t)$
and $(x, y, z) = (5 - 5 t, 13 - 13 t, 7 -7t)$
are parametric equations of the same line.
So you just need some way of limiting the choices of $x_0,$ $y_0,$
and $z_0$ so that you can still make every line by an appropriate choice, but only in one way.
I don't think this needs to be particularly clever;
you just need to reduce the set of points $(x_0,y_0,z_0)$
from three dimensions to two, making sure those two dimensions cover all the lines you want.
Restricting to a plane will get you two dimensions;
the plane does not need to be perpendicular to the lines but it must not be parallel to them.
A: Your approach to (a) will work, but there’s no reason to compute a new pair of basis vectors since you’ve already got one: the normals to the two distinct planes that you’ve already found. Indeed, all of the sought-after planes have equations that are linear combinations of the two equations that you’ve already found.  
It sounds like you did more work than necessary to solve the first part of (a), though. You just need to find two linearly-independent vectors that are perpendicular to $(4,-8,11)$. A simple way to do this is to swap and negate vector entries. That is, given a nonzero vector $(a,b,c)$, its dot products with $(0,c,-b)$, $(-c,0,a)$ and $(b,-a,0)$ are all zero, and at least two of them are nonzero. In this case, you can pick any two of $(0,11,8)$, $(-11,0,4)$ and $(-8,-4,0)$ as the normals to the two planes you’re asked to find. Taking the first and third gives the plane equations $11y+8z=0$ and $2x+y=0$, and per the previous paragraph the one-parameter family of planes $(1-\lambda)(11y+8z)+\lambda(2x+y)=0$ contains all of the perpendicular planes through the origin.  
For part (b) notice that for any two fixed points $(x_0,y_0,z_0)$ and $(x_1,y_1,z_1)$, if their difference is a multiple of the direction vector that you computed, then they generate the same line. To ensure that each line is only listed once, you need a way to generate a set of points that are on distinct lines that share this fixed direction vector. That describes, among other things, a plane perpendicular to these lines, so find a convenient parameterization of such a plane. (The plane doesn’t have to be perpendicular to the lines, but it shouldn’t be parallel to them.)
