Secure function evaluation Consider 3 parties, Alice, Bob and Charlie. Suppose
  each party has a bit as input, i.e. Alice, Bob and Charlie hold $a, b, c \in \{0, 1\}$ respectively.
  Show how construct a scheme with which they can compute the function $f (a, b, c) = a \oplus b \oplus c$
  such that the following are satisfied: 
  
  (1) All parties learn $f(a, b, c)$ at the end. 
  
  (2) No party can learn more about the other party's input than what they can infer from $f(a, b, c)$ and choosing their own input wisely.


My attempt: A chooses $r_1,r_2\in\{0,1\}$, B chooses $s_1,s_2,\in\{0,1\}$, and C chooses $t_1,t_2\in\{0,1\}$ randomly. Then dealer of respective secrets (bold) distribute 3 shares to each party in the following manner.

$A: \it{\bf{a\oplus r_1 \oplus r_2}}, s_1, t_1$

$B: r_1, \it{\bf{b\oplus s_1 \oplus s_2}}, t_2$ 

$C: r_2, s_2, \it{\bf{c\oplus t_1 \oplus t_2}}$.

To reconstruct the secret $a\oplus b\oplus c$, they can compute their sum of shares,

$A: (a\oplus r_1 \oplus r_2)\oplus s_1\oplus t_1$

$B: r_1\oplus( b\oplus s_1 \oplus s_2)\oplus t_2$ 

$C: r_2\oplus s_2\oplus( c\oplus t_1 \oplus t_2)$.
Your protocol is good (assuming an honest-but-curious adversary model). As DrLecter pointed out, each party will need to publish their sum of shares. To recover the answer, each party then simply xors all published values.