(1)  a+2b+3c=12
(2)  2ab+3ac+6bc=48

Solve (1) for 3c

                          3c = 12-a-2b

Write (2) as 

            2ab+a(3c)+2b(3c) = 48

Substitute 12-a-2b for 3c

  2ab+a(12-a-2b)+2b(12-a-2b) = 48

2ab+12a-a^2-2ab+24b-2ab-4b^2 = 48

    -a^2-2ab+12a-4b^2+24b-48 = 0

     a^2+2ab-12a+4b^2-24b+48 = 0

     a^2+2ab-12a+4b^2-24b+48 = 0

    a^2+(2b-12)a+4b^2-24b+48 = 0

Calculate the discriminant B^2-4AC with A=1, B=2b-12, C=4b^2-24b+48

B^2-4AC =
(2b-12)^2-4(1)(4b^2-24b+48) =
(2b-12)^2-4(4b^2-24b+48) =
4b^2-48b+144-16b^2+96b-192 =
-12b^2+48b-48 =
-12(b^2-4b-4) =
-12(b-2)^2
 
In order for this discriminant to be non-negative,
it must be 0

-12(b-2)^2 = 0
   (b-2)^2 = 0
         b = 2

Substitute b=2 in (1)

  a+2b+3c = 12
a+2(2)+3c = 12
   a+4+3c = 12
     a+3c = 8
(3)     a = 8-3c

Substitute b=2 in (1)

          2ab+3ac+6bc = 48
      2a(2)+3ac+6(2)c = 48
(4)        4a+3ac+12c = 48

Substitute 8-3c for a in (4)

  4(8-3c)+3(8-3c)c+12c = 48
  4(8-3c)+3c(8-3c)+12c = 48   
   32-12c+24c-9c^2+12c = 48
          -9c^2+24c+32 = 48
          -9c^2+24c-16 = 0
           9c^2-24c+16 = 0
          (3c-4)(3c-4) = 0
                  3c-4 = 0
                    3c = 4
                     c = 4/3

(3)       a = 8-3c
          a = 8-3(4/3)
          a = 8-4
          a = 4  
  
So a=4, b=2, c=4/3

Answer a+b+c = 4+2+4/3 = 12/3+6/3+4/3 = 22/3

Edwin