There are $\binom{9}{5} = 126$ placements of the $1$’s; subtract those where two or more of the eight lines are constant

Two constant lines must be either a row/column of $1$’s with a parallel line of $0$’s, or two crossing lines of $1$’s, which use exactly five $1$’s

There are $\binom{9}{5} = 126$ matrices satisfying $(1);$ we subtract those with two or more constant lines. Two lines of $0$’s are impossible (they would need at least $5$ zeros), and a line of $1$’s and a line of $0$’s cannot cross, so they must be parallel rows or parallel columns; likewise two lines of $1$’s cannot be parallel ($6$ ones), so they must cross, using exactly $3 + 3 - 1 = 5$ ones.

Case $1:$ a line of $1$’s and a parallel line of $0$’s. There are $6$ choices for the all-$1$ row or column, $2$ for the parallel all-$0$ line, and $3$ ways to fill the remaining parallel line with two $1$’s and one $0:$ $6 \cdot 2 \cdot 3 = 36$ matrices. Every perpendicular line then contains both a $1$ and a $0,$ so no third constant line appears and nothing is double-counted.

Case $2:$ two crossing lines of $1$’s and $0$’s elsewhere. The pair can be a row and a column ($3 \cdot 3 = 9$), a row or column with a diagonal ($6 \cdot 2 = 12$), or the two diagonals ($1$), for $22$ matrices; one checks the four remaining $0$’s never form a constant line. So $126 - 36 - 22 = 68.$