The maximum value is $a^{3}+b^{3}+c^{3}-3abc$. The number of such matrices is $6$. Here is my solution.We know that the absolute value of a determinant does not change with a permutation of the rows and columns.Hence assume WLOG that the first row is is given by $a,b,c$.Also fix the first element of the second row which is eithe $b$ or $c$ as $b$.This forces the determinant to be: $a b c$ $b c a$ $c a b$ which is $3abc-(a^{3}+b^{3}+c^{3})$.As $a+b+c>0$,the result follows. Also it s easy to see that the first row has 6 possible arrangements of $a,b,c$.Also,each of these arrangements gives exactly one determinant with the needed value. Cheers Bhargav