Depth of a union of planes

The ring $A$ embeds in $S\mathrm{/}I\oplus S\mathrm{/}J$ by reduction modulo $I$ and $J$, because $I\cap J=IJ$. Put $\mathrm{\ell }=x+z$. Its image is $(z,x)$ in the two polynomial-domain quotients, so multiplication by $\mathrm{\ell }$ is injective on their direct sum and hence on $A$. Thus $\mathrm{depth}A\ge 1$.

More generally the embedding identifies $A$ with the fibre product of the two local polynomial rings $B=S\mathrm{/}I$ and $C=S\mathrm{/}J$ over their common residue field $k$: its elements are pairs $(f,g)$ with equal constant term. Let $a=(f,g)$ be any regular element of the maximal ideal of $A$. Since $B,C$ are domains, regularity forces $f,g\mathrm{\ne }0$, and both have zero constant term. The pair $(f,0)$ belongs to $A$ but is not in $aA$: an equation $a(h,j)=(f,0)$ would force $h=1,j=0$, while $(1,0)\mathrm{\notin }A$. For every maximal-ideal element $(u,v)$, however, $(u,v)(f,0)=(fu,0)=a(u,0)$ because $(u,0)\in A$. Thus the nonzero class of $(f,0)$ in $A\mathrm{/}aA$ is annihilated by the entire maximal ideal. No regular sequence can extend $a$, so $\mathrm{depth}A\le 1$. Step 1.1 gives the reverse bound. Both irreducible components have dimension $2$, so $\dim A=2$.