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
One of those equations is a condition on $x$, $y$ and $z$ that determines whether the triple $(x,y,z)$ is on the plane. The other is an equation that tells you all the points on the plane as $t_1$ and $t_2$ vary.
(I think you have typos in the second one: you want $$ v = v_0 + t_1v_1 + t_2v_2. $$ )
Here's the analog in two dimensions. You can describe the line through $(1,0)$ and $(2,1)$ either as the set of pairs $(x,y)$ that satisfy $$ x - y -1 = 0 $$ or as the set $$ \{ (1,0) + t(1,1) \ | \ t \in \mathbb{R}\}. $$
You use whichever representation helps you solve the particular problem you are trying to solve.