The largest value of a third order determinant whose elements are 0 or 1, is 

$\text{ Let }\mathrm{\Delta }=\left|\begin{array}{l}{a}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{b}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{c}_1\\ a_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{b}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{c}_2\\ a_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{b}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{c}_3\end{array}\right|$ be a determinant of order 3.

Then,

$\begin{array}{l}\mathrm{\Delta }=a_1{b}_2{c}_3+a_3{b}_1{c}_2+a_2{b}_3{c}_1-a_1{b}_3{c}_2-a_2{b}_1{c}_3-a_3{b}_2{c}_1\\ \mathrm{\Delta }=\left(a_1{b}_2{c}_3+a_3{b}_1{c}_2+a_2{b}_3{c}_1\right)-\left(a_1{b}_3{c}_2+a_2{b}_1{c}_3+a_3{b}_2{c}_1\right)\end{array}$

Since each element of $∆$ is either 1 or 0. Therefore, the value of the determinant cannot exceed 3. Clearly, the value of $∆$ is maximum when the value of each term in first bracket is 1 and the value of each term in the second bracket is zero. But $a_1{b}_2{c}_3=a_3{b}_1{c}_2=a_2{b}_3{c}_1=1$

implies that every element of the determinant is 1 and in that $\mathrm{\Delta }=0$ Thus, we may have

$\mathrm{\Delta }=\left|\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\end{array}\right|=2$