Showing that $\mathcal{V}(f,g) \subseteq \mathbb{A}_k^n$ has three irreducible components. I was working on this following exercise and i would really like to see if i understood things correctly.

Let $V:=\mathcal{V}(X^2-YZ,XZ-X) \subseteq \mathbb{A}_k^n$. Show that $V$ has
three irreducible components and find the corresponding prime ideals.


What i know: I know that any subvariety $V$ of the affine space $\mathbb{A}_k^n$ is a noetherian topological space (with respect to the subspace topology of $\mathbb{A}_k^n$). Now the irreducible components are the maximal irreducible closed subsets of a topological space.
The irreducible components of $V$ correspond to the minimal prime ideals of the coordinate ring $$A(V) = k[X,Y,Z]/{\mathcal{I}(V)}$$ where $\mathcal{I}(V)$ is the Ideal of $V$.

However, the proposed solution to this exercise considers the minimal prime ideals of $k[X,Y,Z]$, which is however the coordinate ring of the affine space $\mathbb{A}_k^n$ and rather than the coordinate ring of $V$.
So I'm guessing, that we are actually looking for the irreducible components of $\mathbb{A}_k^n$ rather than $V$?
What am i missing? Please be aware that I'm new to algebraic geometry and have yet to learn the connections.
My question: Are we supposed to find the minimal prime ideals of the coordinate ring $A(V) = k[X,Y,Z]/{\mathcal{I}(V)}$ or is it actually $A(\mathbb{A}_k^n) = k[X,Y,Z]$ ?
Thanks for any help!
 A: With the help of @AlexWertheim I managed to understand where my confusion was coming from.
An irreducible component $V'$ of $V:=\mathcal{V}(X^2−YZ,XZ−X)\subseteq \mathbb{A}_k^3$ is an irreducible closed subsets of the affine space $\mathbb{A}_k^3$.
Throughout, let ${\frak a}:= (X^2−YZ,XZ−X)$.
Since $V'\subseteq \mathbb{A}_k^3$ is irreducible if and only if its ideal $\mathcal{I}(V') \subseteq k[X,Y,Z]$ is prime, we know that $\mathcal{I}(V')$ must be a prime ideal in $k[X,Y,Z]$.
The irreducible component $V'$ of the affine variety $V$ is a maximal irreducible closed subset of $V$. Therefore, the prime ideal $\frak p$ must be minimal over the ideal of $V =\mathcal{V}({\frak a})$.
The ideal of $\mathcal{V}({\frak a})$ is given by $$\mathcal{I}(\mathcal{V}({\frak a})) = \sqrt{{\frak a}}$$
Note that ${\frak a}\subseteq \sqrt{{\frak a}}$ and $\sqrt{{\frak p}} = {\frak p}$ since prime ideals are radical ideals, it holds that the prime ideals over $\sqrt{a}$ coincide with the prime ideals over $\frak a$ (see comments),  thus it suffices to compute the minimal prime ideals over $\frak a$. Those minimal primes then correspond to the irreducible components of $V$.
In Summary: irreducible components of an affine variety $V\subseteq \mathbb{A}_k^n$ are in particular irreducible closed subsets of $\mathbb{A}_k^n$ and thus correspond to prime ideals in $k[X_1,...,X_n]$. The irreducible components of $V$ being maximal irreducible closed subsets of $V$ then correspond to the minimal primes in $k[X_1,...,X_n]$ over the ideal $\mathcal{I}(V)$ of $V$.
