= Solution
A <prime ideal> $P$ is a proper <ideal> such that $ab\in P$ implies $a\in P$ or $b\in P$. Equivalently $R/P$ is an <integral domain>.
For each $i$, and each $j\ne i$, choose $a_{ji}\in P_j\setminus P_i$, using incomparability. Put
$$
b_i=\prod_{j\ne i}a_{ji}.
$$
Then $b_i$ belongs to the other two <prime ideals> and does not belong to $P_i$, by primality. In particular each $b_i$ lies in the union. But $b_1+b_2+b_3\notin P_i$ for every $i$: modulo $P_i$, the sum equals $b_i\ne0$. \b[The union is not closed under addition and is not an <ideal>.] This is a concrete instance of <prime avoidance>.
The <union of three incomparable nonprime ideals> example can retain pairwise incomparability. In the <ring>
$$
R=\mathbb F_2[u,v]/(u^2,uv,v^2),
$$
take
$$
\boxed{I_1=(u),\qquad I_2=(v),\qquad I_3=(u+v).}
$$
These distinct two-element <ideals> are pairwise incomparable, yet
$$
I_1\cup I_2\cup I_3=\{0,u,v,u+v\}=(u,v),
$$
which is an <ideal>. None is prime: its <quotient ring> still contains a nonzero element of square zero. The three one-dimensional <vector subspaces> cover the two-dimensional <vector space> over $\mathbb F_2$.
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