The general equation of a plane in 3-D is given by $$\mathbf{(p-p_0).n}=0$$
where $\mathbf{p}$ is any general point on the plane, and $\mathbf{p_0}$ is any known point on the plane. $\mathbf{n}$ is a vector normal to the plane.
The equation of a line is given by $$\mathbf{p = l_0+}t\mathbf{l} $$
where $\mathbf{l_0}$ is any point on the line.
If the line lies in the plane, it must satisfy two conditions-
It must be perpendicular to the normal to the plane i.e. $\mathbf{l.n}=0$
$\mathbf{l_0}$ must lie in the plane i.e. satisfy the plane's equation. So, $\mathbf{(l_0-p_0).n} = 0$
You can calculate $\mathbf{p_0,l_0}$ very easily with the information you have.
$\mathbf{p_0}$ can be calculated by choosing any two of $x,y,z$ and finding the third to satisfy the equation of the plane. $\mathbf{l_0}$ is precisely $\vec O$ that you already know. You can use them to cross-check whether the fit is good or not. But they do not depend on $\mathbf{l}$. So, they are secondary, and may be used as a sanity check later on.
Suppose you are given the equation of the line as $ax+by+cx+d = 0$. You can recast it as $[(x,y,z)-\mathbf{p_0}].(a,b,c)$. So, $(a,b,c)$ is your normal vector.
Suppose you have $m$ planes with normals $\mathbf{n_1,n_2,\ldots,n_m}$. Then, your overall constraints are $$\mathbf{l.n_1} = 0 \\ \mathbf{l.n_2} = 0 \\ \ldots \\\mathbf{l.n_m}=0$$.
It is a system of linear equations with 3 variables and $m$ equations. You need to find out $\mathbf{l}$, which can be found up to a constant factor by the standard least squares method, provided $m>3$, which should not be a worry for you. If you code carefully enough, you can implement all of it in matrix terms.
Hope it helps.
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
First we need check that the planes $\Pi_1$ and $\Pi_2$ they are intersects. In order to see it, noticed that $\frac{2}{1}\not=\frac{-1}{-1}$ so $\Pi_{1}\cap\Pi_2=L$ with $L$ the line intersection. Here, we can use two way to find $L$.
Solving the linear system: Indeed, we have $$\begin{cases}x-y-z=1\\2x-y=3\end{cases},$$then $$\begin{pmatrix}1&-1&-1&1\\2&-1&0&3 \end{pmatrix}\sim \begin{pmatrix}1&-1&-1&1\\0&1&2&1 \end{pmatrix}$$ So we have $y+2z=1$ then $y=1-2z$ and $x-y-z=1$ then $x=1+y+z=1+(1-2z)+z=2-z$. Setting $z:=t\in{\bf R}$ we get $$L:\begin{cases}x=2-t\\y=1-2t\\z=t\end{cases},\quad t\in {\bf R}$$
Find the director vector $s$ and a point $P\in L$: The director vector $s$ can be find using the vector product between the normal vectors of the planes, that is,$$s=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&-1&-1\\2&-1&0 \end{vmatrix}=(-1,-2,1)$$ For the point $P\in L$ it is enough to fix one of the variables and solve the system simultaneously for the other variables. Setting $z=0$, we have $$\begin{cases}x-y=1\\2x-y=3\end{cases}$$ Then, $x=2$ and $y=1$ so $P(2,1,0)\in L$ and therefore $$L:\begin{cases}x=2-t\\y=1-2t\\z=0+t\end{cases},\quad t\in {\bf R}$$