Collision resistance of partial SHA-256 hashes XOR'ed
Let tne hashlength be $d$. With $k=12,$ this is the $k-$ XOR problem. Fix $k.$ If the vectors are randomly generated and form a list of size roughly at least $2^{d/k},$ there exists a solution with constant probability bounded away from zero. This is because a list of size $M$ contains $$F:=\binom{M}{k}$$ subsets of size $k,$ and thus as 
$\{x_1,\ldots,x_k\}$ ranges over these subsets the function $f(x_1,\ldots,x_k):=x_1\oplus \cdots \oplus x_k$ ranges over $\{0,1\}^d.$

This is a $F$ balls into $2^d$ bins problem and if $F\geq 2^d,$ the probability that $f$ misses the bin corresponding to your given hash value 
$h_0 \in \{0,1\}^d$ is roughly $e^{-1}\approx 0.37.$ Taking $M=\Omega(k 2^{d/k})$ is enough here. However, finding the solution is computationally more expensive.

Wagner (see here ) has a recursive binary tree based algorithm for the $k-$ XOR problem $$x_1\oplus \cdots \oplus x_k=0 \qquad (1)$$ with $k=2^m,$ with time and memory complexity essentially 
$$O(k 2^{d/(1+m)}).$$ The zero vector on the right hand side of (1) can be replaced by any constant vector.

Let $x_9,\ldots,x_{12},$ be arbitrary and use Wagner with $k=8$ to solve the $8-$ XOR problem

$$x_1\oplus \cdots \oplus x_8=c,$$
with $c=x_9\oplus x_{10} \oplus x_{11} \oplus x_{12}\oplus h_0.$ This can be done with complexity 
$$O(k 2^{d/(m+1)})=O(8 \times 2^{256/4})=O(2^{67}).$$