Thanks Shamrock!
Good hints are So, a[1] is:{{'g', 'h'}, {'i', 'j'}, {'k', 'l'}}
......
Read more last night, and finally I got the answers, just for somebody who hits the same wall as I did.

as (*pa)[2] = &a[1][0], which means *pa is a "2-col" pointer start from {'g', 'h'} in the array a; so that char 'r' is in {'q', 'r'}, the 6th member of pa ---> pa[5], which in turn is a "2-char" array, and with the second index = 1. So the answer is pa[5][1];
as (*ppa)[3][2] = &a[1], which defines *ppa is a "3(row) x 2(col) array " pointer start from {{'g', 'h'}, {'i', 'j'}, {'k', 'l'}} in the array a. For char 'r' is on the 2nd row, 3rd col, within the 3rd col it is in the 2nd col. So the answer is ppa[1][2][1].

The reason why there is multiple answer to get the 'r' is because C actually only has "1-dimension" or linear arrangement of the array members in memory as Corona688 pointed out. When you split the linear array in different combinations, counted by 1, by 2, or by 3, 4, 5.... Actually pa[11] prints 'r', pa[3][2], pa[4][3] pa[5][1]...all point to 'r', but only pa[5][1] is appropriate to the original question. Similar for ppa, but only pa[1][2][1] is the correct answer as it was defined as a 3x2 array.

Shamrock and Corona688, please correct me if I interpreted anything wrong.
Thanks a lot again!

---------- Post updated 09-25-13 at 12:50 PM ---------- Previous update was 09-24-13 at 05:50 PM ----------

To confirm my digestion I tried another two situations by change the pa and ppa definitions:
the answers to get char 'r' are

Code:

pa[7][1]; 
ppa[-1][2][1]

Here it is very interesting the index can be negative. Amazing C!