If ab + bc + ca = 12 and a² + b² + c² = 40, then possible value of \(\frac{1}{2}\) (a + b +c) [(a - b)² + (b - c)² + (c - a)²] is:

224

ab + bc + ca = 12

a² + b² + c² = 40

⇒ (a + b + c)² = [a² + b² + c² + 2(ab + bc + ca)]

                       = 40 + 2 × 12 = 40 + 24 = 64

⇒ (a + b +c) = 8

Now,

⇒ \(\frac{1}{2}\) (a + b +c) [(a - b)² + (b - c)² + (c - a)²]

= \(\frac{1}{2}\) (a + b +c) [2(a² + b² + c²) - 2(ab + bc + ca)]

= \(\frac{1}{2}\) × 8 [(2 × 40) - (2 × 12)]

= 4 (80 - 24) = 224