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You're using bytewise xor, and three plaintexts $x,y,z$ with a small alphabet (A-Z, which I assume means ASCII values 0x41-0x5a). Then knowing $a \oplus x$, at a certain index, we know all possible values of $a$, at that same index, namely $\{0x41 \oplus (a \oplus x),\ldots, 0x5a \oplus (a \oplus x)\}$. But we have 2 more constraints from $a \oplus y$ and $a ...

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