Encryption does not need 7 bit characters - it works on byte data (which is 8 bit). Your Chinese characters will be taken as byte data and encoded (and decoded) correctly without the need for you to play with the data bits.
"I am trying to encrypt each character so that an array of character results a chunk of data as like number of drops results a water.Now we are deviating the actual purpose.Could you please give any idea about what I am asked"
I don't think it is going to help you much, as the "characters" is generates are still not printable, but...
Try:
byte[] data = new byte[3];
Random r = new Random();
r.NextBytes(data);
char[] chars = new char[4];
chars[0] = (char) (data[0] & 0x7f);
chars[1] = (char) (((data[0] & 0x80) >> 7) | ((data[1] & 0x3F) << 1));
chars[2] = (char) (((data[1] & 0xC0) >> 6) | ((data[2] & 0x1F) << 2));
chars[3] = (char) (((data[2] & 0xE0) >> 5));
If your input data is:
b0 abcd efgh
b1 ijkl mnop
b2 qrst uvwx
then this treats it as a least significant bit stream and generates:
c0 bcd efgh
c1 klm nopa
c2 tuv wxij
c3 qrs
You will still probably be better off going with a "proper" encryption algorithm, though...