SOLUTION: There are two triples of positive integers (a,b,c and d,e,f) such that a^2+b^2+c^2 = 86 and d^2+e^2+f^2 = 86. Numbers should not repeat within each triplet. Evaluate the expression

We'll assume a < b < c and d < e < f

a,b,c,d,e,f all have to be 9 or less.  The squares are

1,4,9,16,25,36,49,64,81 

Three of these must have sum 86

We try c = 9

a² + b² + c² = 86
a² + b² + 9² = 86
a² + b² + 81 = 86
a² + b² = 5

We see that the first two, 1 and 4, have sum 5

so (a,b,c) or (d,e,f) can be (1,4,9)

We try c = 8

a² + b² + c² = 86
a² + b² + 8² = 86
a² + b² + 64 = 86
a² + b² = 22

No two of these 1,4,9,16,25,36,49,64,81 add to 22, so

we try c = 7

a² + b² + c² = 86
a² + b² + 7² = 86
a² + b² + 49 = 86
a² + b² = 37

We see that 1+36=37, 

so (a,b,c) or (d,e,f) can be (1,6,7)

Let's let (a,b,c) = (1,4,9) and (d,e,f) = (1,6,7) 

You calculate | abc-def |

Edwin

1)  86 = 1^2 + 2^2 + 9^2


2)  86 = 1^2 + 6^2 + 7^2.


So,  a= 1, b= 2, c= 9;  d= 1, e= 6, f= 7.


     abc = 18,  def = 42,  | abc - def | = 42 - 18 = 24.    ANSWER