Permutation Help required!!

How many even numbers of three digits can be formed with digits 0,1,2,3,4,5 and 6?
Here is my approcah:

The last(unit) digit has to be 0 , 2 , 4 or 6"
So there woeld be 4 cases:

CASE 1:

With '0' as the last(unit digit):

there r 4 choice for the last digit(0,2 , 4 or6) and we select 0,
so there r 6 digits remaining.

there first 2 digits can be selected from the remaining 6 digits in 6p2
ways i.e 30 ways.

So total number if ways for forming 3 digit even number ending with '0' is
4*30 = 120 ways.

CASE 2:

there r 4 choice for the last digit(0,2 , 4 or6) and we select say '2',
so there r 6 digits remaining.

Now, the 10th digit(middle digit), has to be '0' , so there is only one way
for filling the 10th digit. So there r 5 digits remaining

The 100th digit (first digit) can be selected from the reamining 5 digits in
5p1 ways i.e 5 ways

So total number if ways for forming 3 digit even number ending with '2' is
4*5 = 20 ways.

For Case 3 and Case 4: the same logic as Case 2 applies. Since hte 10 digit is '0'.
So for case 3 and 4 , So total number if ways for forming 3 digit even number is 20.

Total number of ways is thus: 120+20+20+20 = 180 ways

Is this approach/answer rite?

The question was posted in a website and there were many answere(120, 105, 168 etc) not sure which is the rite answer?

Anu