Have a look at the wikipedia article. They show a 15 bit signal I think. Just consider the leftmost 7 bits because that's what you have.
You will notice that there are parity bits spread through the data. In your case it's the first, second, and fourth bits.
The first bit contains the parity of bit 1, 3, 5, and 7
The second bit contains the parity of bits 2, 3, 6, and 7
The fourth bit contains the parity of bits 4, 5, 6, and 7.
(If you write the bit numbers down in binary, you can see that each parity bit corresponds to locations where the corresponding bit is 1)
321
001 = 1 (P1)
010 = 2 (P2)
011 = 3 (D1) (covered by P1 and P2)
100 = 4 (P3)
101 = 5 (D2) (covered by P1 and P3)
110 = 6 (D3) (covered by P2 and P3)
111 = 7 (D4) (covered by P1, P2, and P3)
Notice that I have numbered the bit positions in binary and labelled the columns from LSB to MSB.
Where Column 1 is 1, that bit is covered by P1
Where Column 2 is 1, that bit is covered by P2
Where Column 3 is 1, that bit is covered by P3
P1, P2, and P3 are in the position where *ONLY* the corresponding bit is 1.
So all you need to do is write down all the bits next to this, then calculate what the parity bits should be. If they're all right, the message is assumed to be OK. If they're not OK, you need to flip a single bit to fix them.
If only 1 parity bit is wrong, it's the parity bit that needs flipping.
If 2 parity bits are wrong, you need to flip the data bit that is covered ONLY by those 2 parity bits.
Both rules can be generalised to cover any situation by saying that the bit which needs to be flipped is the one covered by ONLY the parity bits which are wrong.
If you don't know whether the parity is odd or even, there are 2 possible solutions to each problem.
Does that help?