The correct answer is option (B):
32
We are given that a + b + c = 11 and ab + bc + ca = 35. We want to find the value of (a-b)^2 + (b-c)^2 + (c-a)^2.
Let's expand the expression we want to find:
(a-b)^2 + (b-c)^2 + (c-a)^2 = (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ca + a^2)
= 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca
= 2(a^2 + b^2 + c^2) - 2(ab + bc + ca)
We know that (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).
We are given a+b+c=11, so (a+b+c)^2 = 11^2 = 121.
Also, we are given ab + bc + ca = 35.
Therefore, 121 = a^2 + b^2 + c^2 + 2(35)
121 = a^2 + b^2 + c^2 + 70
a^2 + b^2 + c^2 = 121 - 70 = 51
Now, we can substitute this value back into our expanded expression:
2(a^2 + b^2 + c^2) - 2(ab + bc + ca) = 2(51) - 2(35)
= 102 - 70 = 32
Therefore, (a-b)^2 + (b-c)^2 + (c-a)^2 = 32.