Since A,B,C are consecutive digits,

B=A+1, C=A+2

Let the digits of the third number be 

A+x, A+y and A+z where {x,y,z} = {0,1,2}

[[100A+10(A+1)+(A+2)]+[100(A+2)+10(A+1)+A]+[100(A+x)+10(A+y)+(A+z) = 1242

That simplifies to

333A+100x+10y+z = 1020

     100x+10y+z = 1020-333A

The left side is the number "xyz"
and it is a 2 or 3 digit number with 
digits {0,1,2} in some order.

The least it could be is 012, or 12, and the most it could be is 210

So    

Subtract 1020 from all three sides:

      

Divide all three sides by -333

       
 
The only integer between those values is A = 3

Thus A = 3

Substituting in

100x+10y+z = 1020-333A
100x+10y+z = 1020-333(3)
100x+10y+z = 1020-999
100x+10y+z = 21

So x must be 0 otherwise the left side will be more than 21

100(0)+10y+z = 21
       10y+z = 21

So the only way that can be true is for y=2  and z=1

So the digits in the third row are

A+x, A+y, A+z or 3+0, 3+2, 3+1 or 3, 5, 4

   A B C                  3 4 5
+  C B A               +  5 4 3
+  - - -               +  3 5 4
--------------         --------------
 1 2 4 2                1 2 4 2
--------------         --------------

Edwin