Are there different types of zero vectors? Take the cross product
$\vec{a}×\vec{b}=0\tag{1}$
Where $|\vec{a}|,$$|\vec{b}|$$≠0$ and the angle between $\vec{a}$ and $\vec{b}$ is $0$
Cross product of two vectors can be used to find the vector normal to those two vectors.
For example:
$\vec{x}×\vec{y}= $$|\vec{x}||\vec{y}|\sin\theta\hat{n}$
$\hat{n}$ being normal unit vector to $\vec{x}$ and $\vec{y}$
So, in $(1)$, isn't the answer $0\hat{n}$? $\hat{n}$ being normal to both $\vec{a}$ and $\vec{b}$, i.e., lying in a plane to which both $\vec{a}$ and $\vec{b}$ are normal?
I know that a $\vec{0}$ has an arbitrary direction, but in the eq$(1)$, the $0\hat{n}$ vector has arbitrary direction in one plane. So it's "arbitrariness" is limited to a particular plane.
Is the above argument, that there are different types of zero vectors, correct?
If not, What is the flaw? And What exactly is a zero vector? What are its applications?
 A: The defining property of the zero  vector is that given some other vector $v$ you have $0 + v = v$. To see  that the zero vector is unique we note that if $0_a$ and $0_b$ both satisfy this relation then $0_a=0_a+0_b=0_b$ so it is unique.
We'll also want to check that $0v=0$ with the $0$ on the left being a scalar. To do this we calculate $0v=(0+0)v=0v + 0v$ and so adding $-0v$ to both sides we have $0v=0$ as required.
You have stumbled upon the fact that the $0$ vector is orthogonal to every other vector. This proves problematic when trying to assign a direction to the $0$ vector which is why it's typically considered directionless.
I would add that in terms of applications you're often interested in linear maps, and if we have some linear map $\varphi$ then the preimage of the $0$ vector $\varphi^{-1}(0)$ will be a vector space. This space is so important that it gets its own name, the nullspace or the kernel of the transformation. So algebraically its the most important vector.
As for applications you can make certain assumptions about a system by examining how you expect the nullspace to behave. For example you'd expect a change of basis or a rotation to keep the nullspace trivial only mapping the $0$ to $0$. However for an orthogonal projection operator you'd expect a non-trivial nullspace since there is a loss of dimension. Sometimes the nullspace will be a set of valid or optimal solutions to an optimization problem as well so you can characterize them by examining the nullspace.
A: There are different ways to see the vectorial product. The first is the one you gave, which is the angle between the two vectors $\vec a$ and $\vec b$. Consider two vectors $\vec u$ and $\vec v$ on the $Oxy$ plane such as $\vec u \times \vec v =: \vec w$:

You see that the more they are parallel, the less is the norm of $\vec w$. You also see that $\vec w$ changes its direction regarding the quadrants of the two vectors $\vec u$ and $\vec v$: this coresponds to the behaviour of $\sin \theta$. So the fact that $\vec w = \vec 0$ only express that $\sin 0 = 0$ here.
A second way to see the vectorial product, is that it shows the normal vector to a plane:

Here, the plane is defined by the vectors $\vec u$ and $\vec v$ linearly independent. $\vec w$ shows the normal direction to the plane, so if $\vec w = (w_x, w_y, w_z)$, the equation of the plane is:
$$w_x x + w_y y + w_z z = 0$$
If $\vec w = 0$, it shows that you have taken two linearly dependent vectors, and that they are simply not enough information to describe the plane. You get $0 = 0$, which is kind of an "error message" showing you need to take a different $\vec v$. So, again, their is nothing "special" about this zero vector.
A last good example is the curl:

It is also defined as a vectorial product and describes the rotating behaviour of a force field. Here, $\vec w = \vec 0$ expresses the fact that the field is laminar, so their is no rotating effect.
To conclude, I don't think we can talk about "different types of zeros" here. For me, the zero solution only expresses a particular case of the studied phenomenon. If I didn't understand something you meant, don't hesitate to point it out.

Note that the second answer to your question confirms my thoughts:

Hence all Zero Vectors are Exactly Same.

A: Consider $\vec{a}×\vec{b}=\vec{c}$
Here , $\vec{c}$ may have "unique" Direction Perpendicular to the Plane in which $\vec{a}$ & $\vec{b}$ lie.
Direction is "unique" when $|a|$ & $|b|$ & $\sin(\theta)$ ($\theta$ is the angle between the two vectors) are all not $0$.
When $\vec{c}=|a||b|\sin(\theta)\hat{n}=0\hat{n}$ , you are trying to claim that the Direction is unique , hence there are many "Zero Vectors" , unfortunately , that is wrong because when it is Zero Vector, the Direction is meaningless , not unique.

Consider these 3 subcases :
$|a|=0$ OR $|b|=0$ : there is no unique Plane containing $\vec{a}$ & $\vec{b}$ , we have a line & a Point here.
$|a|=0$ AND $|b|=0$ : there is no unique Plane containing $\vec{a}$ & $\vec{b}$ , we have 2 Points here.
$\theta=0$ : there is no unique Plane containing $\vec{a}$ & $\vec{b}$ , we have 2 overlapping lines here.

When the Plane is not available , we have no Direction normal to that Plane.
Hence Zero Vector has no unique Direction.
Hence all Zero Vectors are Exactly Same.

When you state :
[[ I know that a $0\hat{n}$  has an arbitrary direction, but in the eq(1), the $0\hat{n}$ vector has arbitrary direction in one plane. So it's "arbitariness" is limited to a particular plane.]]
There is no Particular Plane , that is where your logic is going wrong.
