The number of 6 digit number that can be formed using the digits 0,...

Only one possibility is there, $( 9,1, 2 )$ or $( 0,5, 7 )$ as in even or odd places
$∴$ Total number of required $6$ digit numbers

$= ( 9, 1, 2$~in odd and& $0, 5, 7$~in even places$)+( 9, 1, 2$~in even places and& $0, 5, 7$~in odd places$)$
$= 3 ! \times 3 ! + \left(3 ! – 2\right) \times 3 !$
$= 36 + 24$
$= 60$