If
$v = (4, 0 , -5) \tag{1}$
then clearly
$w_1 = (0, 1, 0) \tag{2}$
is orthogonal to $v$; so is
$w_2 = (5, 0, 4), \tag{3}$
so $w_1$ and $w_2$ must span the plane normal to $v$; I found $w_1$ and $w_2$ by simple inspection, that is, intelligent (I hope!) guesswork! The parametric equation of the plane is then
$(x, y, z) = sw_1 + tw_2 = s(0, 1, 0) + t(5, 0, 4) = (5t, s, 4t). \tag{4}$
We might also observe that $w_1$ is normal to $w_2$, so if we normalize $w_2$ to
$w_2' = (\dfrac{5}{\sqrt{41}}, 0, \dfrac{4}{\sqrt{41}}) \tag{5}$
we can express the plane in terms of the orthonormal pair $w_1$, $w_2'$:
$(x, y, z) = sw_1 + tw_2' = (\dfrac{5}{\sqrt{41}}t, s, \dfrac{4}{\sqrt{41}}t); \tag{6}$
and finally, we can always use the non-parametric vector form
$0 = v \cdot (x, y, z) = 4x - 5z. \tag{7}$
The vector $v$ given by (1) may be normalized to
$v' = (\dfrac{4}{\sqrt{41}}, 0, -\dfrac{5}{\sqrt{41}}) \tag{8}$
if so desired; then (7) becomes
$0 = v' \cdot (x, y, z) = \dfrac{4}{\sqrt{41}}x - \dfrac{5}{\sqrt{41}}z. \tag{9}$
Note that in the above I have been able to neglect inclusion of the point $P_0$ since here, as in the OP's $2$-dimensional example, $P_0 = 0$, the zero vector.
Hope this helps. Cheers,
and as always,
Fiat Lux!!!
Actually, the method that you’re trying to apply would not produce the correct answers for single point. The method finds the tangent plane to a surface at a point on the surface, but neither of the given points lies on the surface (well, $A$ would be on the hyperboloid if there weren’t a slice missing along the plane $x=0$). It makes no sense to compute $\vec n$ at these points. Instead, what you need to do is construct the equation of the tangent plane at the generic point $P_0=(x_0,y_0,z_0)$ on the surface and then plug in the coordinates of the two known points to solve for $P_0$. You’ve already worked out that $$\vec n_{P_0} = \left[{1-y_0^2\over x_0^2},2{y_0\over x_0},-1\right]$$ so the equation of the unknown plane is $\vec n_{P_0}\cdot(P-P_0)=0$. Substitute the coordinates of $A$ and $B$ for $P$ and solve the resulting system of equations for $P_0$.
Since you’ve tagged this question with “linear algebra,” here’s a linear-algebraic way to solve this. Rewrite the equation of the surface as $xz=y^2-1$ and keep in mind that the domain excludes $x=0$. This is the equation of a quadric surface which can be written in matrix form as $$\mathbf x^TQ\mathbf x = \begin{bmatrix}x&y&z&1\end{bmatrix} \begin{bmatrix} 0&0&-\frac12&0 \\ 0&1&0&0 \\ -\frac12&0&0&0 \\ 0&0&0&-1 \end{bmatrix} \begin{bmatrix}x\\y\\z\\1\end{bmatrix} = 0.$$ The equation $ax+by+cz+d=0$ of a plane can also be written as $(a,b,c,d)^T(x,y,z,1)=0$, so this plane can be represented by the vector $\mathbf\pi = (a,b,c,d)^T$. It turns out that all of the planes tangent to a nondegenerate quadric $Q$ satisfy the dual equation $\mathbf\pi^TQ^{-1}\mathbf\pi=0$ and that if $\mathbf\pi$ is tangent to $Q$, then its point of tangency is $\mathbf p = Q^{-1}\mathbf\pi$. (These equations can be derived from pole-polar relationships.)
Now, by substituting the coordinates of $A$ and $B$ into the generic plane equation, we see that the coefficients in the equations of any plane that passes through these points satisfies the system $$b+d=0 \\ a+3b+4c+d=0,$$ i.e., the representative vectors of this family of planes are the null space of $$\begin{bmatrix}0&1&0&1\\1&3&4&1\end{bmatrix}.$$ Since multiplying both sides of an equation by a nonzero constant doesn’t change the solution set, vectors that are multiples of each other represent the same plane, so to describe the family of planes through $A$ and $B$ it suffices to consider only convex combinations of basis vectors for this null space. Computing a basis for the null space via row-reduction produces $\mathbf\pi(\lambda) = (2-6\lambda,\lambda-1,\lambda,1-\lambda)^T$. So, finding the tangent planes through $A$ and $B$ becomes a matter of solving $\mathbf\pi(\lambda)^TQ^{-1}\mathbf\pi(\lambda)=0$ for $\lambda$. For this particular problem, inverting $Q$ is quite simple. This equation is quadratic in $\lambda$, so there might be two solutions. If you check where they are tangent to the surface, however, you’ll find that one of the points of tangency has the forbidden $x=0$, so you must reject that potential solution.
Best Answer
According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:
The normal vector then would be$$v=v_1\times v_2=\begin{bmatrix}1\\2\\1\end{bmatrix}\times \begin{bmatrix}2\\-4\\1\end{bmatrix}=\begin{bmatrix}2\times 1-1\times(-4)\\2\times 1-1\times 1\\1\times(-4)-2\times 2\end{bmatrix}=\begin{bmatrix}6\\1\\-8\end{bmatrix}$$therefore $$(x,y,z)\cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$