A symmetric master key is distributed between two data centres by courier in three parts: $T_1 = \mathtt{6C}$, $T_2 = \mathtt{15}$, $T_3 = \mathtt{A9}$, each in a sealed envelope, and both sides compute $K = T_1 \oplus T_2 \oplus T_3$. What is $K$, and what does a courier learn who secretly opens one envelope (or even two)?

Question

Accepted Answer

$K = \mathtt{D0}$. A courier with one or two parts learns nothing about $K$, even with unlimited computing power, as long as one part stays unknown.
Computing the key (hex digits, bitwise XOR):

$\mathtt{6C} \oplus \mathtt{15} = \mathtt{79}$
$\mathtt{79} \oplus \mathtt{A9} = \mathtt{D0}$

Why the leaked parts are worthless: every part is uniformly random. The one part the attacker does not know acts exactly like a one-time pad over the XOR of the others: for every possible key value there is exactly one value of the missing part that would produce it. So the attacker knows no more than before, namely that $K$ lies somewhere between $\mathtt{00}$ and $\mathtt{FF}$. This is information-theoretic security, not just computational.
Why the signed receipts matter: the courier signs that each envelope arrived undamaged, so tampering would at least be noticed. Splitting into $n$ XOR parts means an attacker must obtain all $n$.