Correct Answer - Option 1 :

\(- \dfrac{1}{2} \ and \ 1\)
**Given :**

The sum of Squares of 3 distinct numbers = 1

The sum of their product taken two at a time , lies in between

**Calculation :**

Let a, b and c be three numbers such that:

a^{2}+b^{2}+c^{2 }= 1_____(1)

Also,

we are asked to find the range in which (ab +bc + ca) lies.

Clearly from equation (1) we could observe that:

a ,b, c lie between -1 and 1

hence ,a.b

Also, when a=1 and b=c=0 (Then equation (1) is satisfied)

then, ab + bc + ca = 0

Hence, option: (B) and (C) are discarded.

Also, we know that:

⇒ (a + b + c)^{2} = a^{2 }+ b^{2} + c^{2 }+ 2ab + 2bc

As (a + b + c)^{2} > 0

Hence,

⇒1 + 2(ab + bc + ca) > 0

⇒(ab + bc + ca) > -1/2

Hence,

⇒ -1/2 <ab + bc + ca <1

**∴ The required answer is -1/2 and 1**