Assume   x@y = Ax + By + C

Substitute x=7, y=8

         7@8 = A(7) + B(8) + C
          31 = 7A + 8B + C
 7A + 8B + C = 31

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Substitute x=7, y=8

         4@6 = A(4) + B(6) + C
          21 = 4A + 6B + C
 4A + 6B + C = 21

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Substitute  x=5, y=12

        5@12 = A(5) + B(12) + C
          35 = 5A + 12B + C
5A + 12B + C = 35

So we have this system of equations:

7A +  8B + C = 31
4A +  6B + C = 21
5A + 12B + C = 35

Solve that system and get A=2, B=2, and C=1

Substitute in

   x@y = Ax + By + C

   x@y = 2x + 2y + 1

Therefore to find 3@2 we substitute x=3 and y=2

   3@2 = 2(3) + 2(2) + 1
   3@2 = 6 + 4 + 1
   3@2 = 11

So the correct choice is B.

Edwin