Let $$Det(P) = \left| {\matrix{ {{a_2}} & {{b_2}} & {{c_3}} \cr {{a_3}} & {{b_3}} & {{c_3}} \cr } } \right|$$

$$ = {a_1}({b_2}{c_3} - {b_3}{c_2}) - {a_2}({b_1}{c_3} - {b_3}{c_1}) + {a_3}({b_1}{c_2} - {b_2}{c_1})$$

Now, maximum value of Det (P) = 6

If $${a_1} = 1$$, $${a_2} = - 1$$, $${a_3} = 1$$, $${b_2}{c_3} = {b_1}{c_3} = {b_1}{c_2} = 1$$ and $${b_3}{c_2} = {b_3}{c_1} = {b_2}{c_1} = - 1$$

But it is not possible as

$$({b_2}{c_3})({b_3}{c_1})({b_1}{c_2})$$ = $$-$$1

and $$({b_1}{c_3})({b_3}{c_2})({b_2}{c_1})$$ = 1

i.e., $${b_1}{c_2}{b_3}{c_1}{c_2}{c_3}$$ = 1 and $$-$$1

Similar contradiction occurs when

$${a_1} = 1$$, $${a_2} = 1$$, $${a_3} = 1$$, $${b_2}{c_1} = {b_3}{c_1} = {b_1}{c_2}$$ = 1 and $${b_3}{c_2} = {b_1}{c_3} = {b_1}{c_2} = - 1$$

Now, for value to be 5 one of the terms must be zero but that will make 2 terms zero which means answer cannot be 5

Now,

$$\left| {\matrix{ 1 & 1 & 1 \cr { - 1} & 1 & 1 \cr 1 & { - 1} & 1 \cr } } \right| = 4$$

Hence, maximum value is 4