if 2s = a + b + c then what is: 

s(s - a)(s - b)(s - c) \(\left[ \frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s} \right] \)

Given:

2s = a + b + c ⇒ s = (a + b + c)/2

s(s − a)(s − b)(s − c) × \(\left[\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s}\right]\)

Let us take:

a = b = c = 2 ⇒ a + b + c = 6 ⇒ 2s = 6 ⇒ s = 3

Then: s − a = s − b = s − c = 1

Now evaluate:

s(s − a)(s − b)(s − c) × \(\left[\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s}\right]\)

= 3 × 1 × 1 × 1 × [1/1 + 1/1 + 1/1 − 1/3]

= 3 × (3 − 1/3) = 3 × (8/3) = 8

Check which option gives 8:

abc = 2 × 2 × 2 = 8 (Match)

2abc = 16 (Not matching)

4abc = 32 (Not matching)

ab + bc + ca = 12 (Not matching)

Therefore, the correct answer is option 1.