No of keys used in Triple DES Algorithm I am studying Triple DES.

I have got two basic questions. One, why are we following a sequence like encrypt(decrypt(encrypt)); what is the problem using a sequence encrypt(encrypt(encrypt))? 

Second, why are we not using a single long key (say 168 independent key bits) in lieu of three individual 56 bits keys? 

Thanks.
The first question is already answered here.

For the second question, a single key $k \in \left\{0,1\right\}^{168}$ can be trivially converted back and forth from three keys $k_1, k_2, k_3 \in \left\{0,1\right\}^{56}$ – there is no semantic difference between the two. DES itself operates only on 56-bit keys, and triple-DES requires three independent DES operations. Software that implements triple-DES can trivially be written to either accept a single 168-bit key or three 56-bit keys.