Option : (A)
f(x) = (x - a)2 +(x - b)2 +(x - c)2
f'(x) = 2[x – a + x – b + x - c]
f'(x) = 2[3x - a - b - c]
f’(x) = 0
2[3x - a - b - c] = 0
3x - a - b- c = 0
3x = a + b + c
x = \(\frac{a+b+c}{3}\)
Hence by second derivative test f’’(x) > 0 so it’s a point of minimum.
f” (x) = 2(3(1))
= 6 > 0
So,
x = \(\frac{a+b+c}{3}\) point of minimum.