The easiest way here is to determine $a$ such that the line directional vector and the plane normal vector are perpendicular. For this, their dot product needs to be zero, i.e.
$$(a,1,2) \cdot (a,a,-1) = 0$$
(The coordinates of the plane normal vector are the coefficients next to $x$, $y$ and $z$ respectively in the implicit equation of the plane.)
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.)
Best Answer
The line of intersection is in both planes, not perpendicular to them. You have two equations in three unknowns, which should have a solution depending on one parameter, just like a line. So can you solve the two equations to give something like that?
Added: If you subtract the two equations, you get y=1. If you plug that into the first, they become identical, x+z=0. So all points on both planes have to satisfy both of these. A parameterization of the line would then be (t,1,-t)