Let the 3 consecutive integers be n-1,n,n+1. 

Now 2(n+1) = 3(n-1) - 3, ie., 3n = 6 + 2n +2, ie., n = 8.

Then numbers are 7,8,9.

Integers are x , x  + 1 , x + 2

2 ( x  + 2 ) = 3x  - 3

2x  + 4  = 3x  - 3

7   = x

Integers are 7 , 8 , 9

7, 8 and 9 are the three consecutive integers as required.

Three consecutive integers such that,   2(x + 1) = 3(x - 1) - 3   ,

twice the greatest integer is 3 less than 3 times the least integer:

7, 8, and 9.

n – 1, n, n + 1 

2(n + 1) = 3(n – 1) - 3 

2n + 2 = 3n – 6 

Middle number n was 8, so, 7, 8, 9 

 Three consecutive integers:

 x, (x  + 1), (x + 2)

Twice the greatest integer is 3 less than 3 times the least integer

2(x + 2) = 3x - 3

2x + 4 = 3x - 3

x = 7

x + 1 = 8

x + 2 = 9

ANS 7, 8 and 9

Check:

2 x 9 = 18

3 x 7 = 21

21 - 3 = 18

Three consecutive integers.

I'll call the middle number "x", so the one before and after it are: (x - 1) and (x + 1)

Twice the greatest: 2(x + 1)

is 3 less than 3 times the least: 3(x - 1) - 3

Set them equal and solve for x:

2(x + 1) = 3(x - 1) - 2

2x + 2 = 3x - 3 - 2

2x + 2 = 3x - 5

-x = -7

x = 7

Remember that I called "x" your middle number, so your three numbers are:

6, 7, and 8