XOR cipher with fixed key and known relation among plaintexts
I'm somewhat new at this, so there might be a better way to solve this, but this is how I solved it. If I understand right, the extra parameter given is written as:
$$\mathtt{}({e}_{1} \oplus k) \oplus ({e}_{2} \oplus k) = e_{3} \oplus k$$
(that is, the decoded contents of e1 xor'd with the decoded contents of e2 is equal to the decoded contents of e3)
The brackets only provide readability, and so the equation is the same without them. This means that we have $$\mathtt{}{e}_{1} \oplus k \oplus {e}_{2} \oplus k = e_{3} \oplus k$$
There are 2 &quot;$\mathtt{}\oplus k$&quot;'s on the left hand side here, so they can be removed to get the following:
$$\mathtt{}{e}_{1}\oplus {e}_{2} = e_{3} \oplus k$$
The right hand side can also be replaced with $\mathtt{}m_{3}$ to denote decrypted $\mathtt{}e_{3}$:
$$\mathtt{}{e}_{1}\oplus {e}_{2} = m_{3}$$
$\mathtt{}e_{1}$ and $\mathtt{}e_{2}$ are given, and by xoring them we can get:
$$\mathtt{}{m}_{3} = 1101001100$$
and we also know that:
$$\mathtt{}{m}_{3} = {e}_{3} \oplus {k}$$
Now we have both $\mathtt{}m_{3}$ and $\mathtt{}e_{3}$, and we can now solve for $\mathtt{}k$:

xor both sides by $\mathtt{}e_{3}$
$$\mathtt{}{m}_{3} \oplus {e}_{3} = {e}_{3} \oplus {e}_{3} \oplus {k}$$
remove $\mathtt{}e_{3} \oplus e_{3}$ from right hand side:
$$\mathtt{}{m}_{3} \oplus {e}_{3} = {k}$$
substitute in $\mathtt{}m_{3}$ and $\mathtt{}e_{3}$'s values:
$$\mathtt{}1101001100 \oplus 11010110101 = {k}$$
giving...:
$$\mathtt{}k = 10111111001$$

You can now decode all 3 encrypted messages:
$$\mathtt{}m_{1} = 10011000011$$
$$\mathtt{}m_{2} = 11110001111$$
$$\mathtt{}m_{3} = 01101001100$$
Hope that helps.