The number of matrices of order 3 × 3 , whose entries are either 0...

Let the matrix be \left[\begin{cases}a & b & c \\ d & e & f \\ g & h & i\end{cases}\right]

Since the elements can be either of $0\text{or}1$, so

$a + b + c + d + e + f + g + h + i = 2 / 3 / 5 / 7$

For sum as $2$, the number of cases will be $\dfrac{9 !}{2 ! 7 !} = 36$

For sum as $3$, the number of cases will be $\dfrac{9 !}{3 ! 6 !} = 84$

For sum as $5$, the number of cases will be $\dfrac{9 !}{5 ! 4 !} = 126$

For sum as $7$, the number of cases will be $\dfrac{9 !}{2 ! 7 !} = 36$

Total required matrix will be $282$