We have an even 3 digits number whose sum lie is less than 9, has got 3 digits and less than 140.
We will establish the inequalities that satisfies the conditions given and then figure out the number.
![\begin{gathered} 100x+10y+z<140 \\ x+y+z<9 \\ 100x+10y+z=14a\text{ where a lies between 8 and 9} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/izb9nzlf0o1lhkwcqjxdgkxzc7ne28ha3m.png)
From our last inequality, we can easily see that the number in question is 14 x 8 or 14 x 9. Any multiple of 7 that is even is also a multiple of 14.
![\begin{gathered} 14*8=112\text{ AND} \\ 14*9=126 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2oauac4612c4dwi1qxpxmdlh707vro4pzf.png)
From the above, it can be easily seen that 112 satisfies the conditions listed.
The number is 112