Let P be a matrix of order 3 × 3 such that all the entries in P are...

∆ = \left|$\begin{cases}{a}_{1} & {a}_{2} & {a}_{3} \\ {b}_{1} & {b}_{2} & {b}_{3} \\ {c}_{1} & {c}_{2} & {c}_{3}\end{cases}$\right| = $$\begin{array}{ll}\mathop{\left({a}_{1}{b}_{2}{c}_{3} + {a}_{2}{b}_{3}{c}_{1} + {a}_{3}{b}_{1}{c}_{2}\right)}_{⏟} \\ x\end{array}$$ - $$\begin{array}{ll}\mathop{\left({a}_{3}{b}_{2}{c}_{1} + {a}_{2}{b}_{1}{c}_{3} + {a}_{1}{b}_{3}{c}_{2}\right)}_{⏟} \\ y\end{array}$$$<br />Now if$x \le 3$and$y \ge - 3$<br />The$∆$can be maximum 6 but it is not possible<br />because$x = 3 ⇒$each term of$x = 1$<br />And$y = -3 ⇒$each term of$y = - 1$<br />$⇒ \mathop{∏}_{i = 1}^{3}{a}_{i}{b}_{i}{c}_{i} = 1$and$\mathop{∏}_{i = 1}^{3}{a}_{i}{b}_{i}{c}_{i} = - 1$, which is a contradiction.<br />Now for value to be$5$one the terms must be zero but that will make two terms zero which means answer cannot be$5$. Next possibility is$4which can be obtained as \left|\begin{cases}1 & 1 & 1 \\ - 1 & 1 & 1 \\ 1 & - 1 & 1\end{cases}\right| = 1\left(1 + 1\right) - 1\left(- 1 - 1\right) + 1\left(1 - 1\right) = 4