How hard is it to find plaintexts whose hashes satisfy $h(a)\oplus h(b)=h(c)$? Given a cryptographic hash function $h$, for example SHA256, how hard is it to find plaintexts $a,b,c$ such that
$$h(a)\oplus h(b)=h(c) \text?$$
This partial answer establishes (rather trivial) lower and upper bounds for the asymptotic hardness of the problem, assuming $h$ behaves like an $n$-bit wide random function.
If one hashes $m$ messages $M$, then computes $f(i,j,k)=h(M_i)\oplus h(M_j)\oplus h(M_k)$ for $(i,j,k)\in\mathbb {Z_m}^3$, that's $m^3$ results, with most values duplicated at least 6 times for large $m$. Odds that zero is never reached by any $f(i,j,k)$ are about $(1-2^{-n})^{m^3/6}$ for large $m$, for any choice of the $M$ by an adversary who can't distinguish $h$ from a random function.
Therefore, if $m\approx2^{n/3}$ hashes have been computed, that can not lead to a solution of the problem with odds better than 16% for large parameters.
The expected hardness of the problem in the question is thus at least $\mathcal O(2^{n/3})$ times the work for a hash. Also, an infinitely powerful adversary could succeed with $\mathcal O(2^{n/3})$ queries to an oracle implementing the hash.

The above leads to an explicit algorithm with non-vanishing odds of success that performs $(2^{n/3})^3=2^n$ evaluations of $f$, thus has cost $\mathcal O(2^n)$ $n$-bit operations; it requires $\mathcal O(2^{n/3})$ $n$-bit words of memory. We can do much better.
One option is to fix $a$, then find collisions for the function
$$g(x)=\begin{cases}h(x) &amp; \text{if }x\text{ is even}\\h(x)\oplus h(a) &amp; \text{if }x\text{ is odd}\end{cases}$$
This search can be can be made with cost $\mathcal O(2^{n/2})$ hashes and modest memory, using Floyd's cycle-finding, or Paul C. van Oorschot and Michael J. Wiener's Parallel Collision Search with Cryptanalytic Applications (in Journal of Cryptology, January 1999, Volume 12, Issue 1; free slightly earlier version available from the first author's website).
Notice that if $g(b)=g(c)$ with $b\ne c$, then $(a,b,c)$ is a solution of the problem of the question if $b$ and $c$ have different parity, which (for random $h$) has odds about $1/2$ for each collision exhibited.
The expected hardness of the problem in the question is thus at most $\mathcal O(2^{n/2})$ times the work for a hash.