= Solution
Any isomorphism would permute <irreducible components> and preserve their intersections. In each configuration, the origin is the unique point lying on all three components, so an isomorphism must take origin to origin.
The <coordinate ring> of the spatial configuration is
$$
A=k[x,y,z]/(xy,xz,yz).
$$
The displayed ideal is indeed the intersection of the three coordinate-axis ideals. Its generators have no linear parts, so at the origin the residue classes of $x,y,z$ form a basis of $\mathfrak m/\mathfrak m^2$. Hence its <algebraic cotangent space> and <Zariski tangent space> have dimension three.
For the plane configuration, the <coordinate ring> is $B=k[s,t]/(st(s-t))$. The three linear factors are distinct in every characteristic, so this principal ideal is radical. It likewise has no linear part, but now $\mathfrak n/\mathfrak n^2$ has basis $s,t$ and dimension two. An isomorphism of <local rings> induces an isomorphism of these cotangent spaces, which is impossible:
$$
\boxed{\operatorname{edim}\mathcal O_{X,0}=3\ne2=\operatorname{edim}\mathcal O_{Y,0}.}
$$
Thus \b[the varieties are not isomorphic]. This is <embedding dimension distinguishes three-branch curves>: matching incidence of branches does not determine their local algebra.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-21-three-branches.png]
{title=Three spatial axes and three coplanar lines have different tangent-space dimensions at their common point}
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