1. We are given

   \qquad a,b,c\text{ are distinct,} \qquad a^2+b^2+c^2=1.$$ We need to determine the value of $$ab+bc+ca.$$

2. Use the identity $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).$ Also, $(a-b)^2+(b-c)^2+(c-a)^2 \ge 0.$

3. Expanding this,

   =2(a^2+b^2+c^2)-2(ab+bc+ca). $$ Hence, $$2(a^2+b^2+c^2)-2(ab+bc+ca)\ge 0,$$ so $$a^2+b^2+c^2 \ge ab+bc+ca.$$

4. Since $a^2+b^2+c^2=1$, we get $ab+bc+ca \le 1.$

5. But $a,b,c$ are distinct, so equality cannot hold. Equality in $a^2+b^2+c^2 \ge ab+bc+ca$ occurs only when $a=b=c,$ which is impossible because they are distinct.

6. Therefore, $ab+bc+ca<1.$

7. Checking options:

   • A: less than 1 — correct
   • B: equal to 1 — false
   • C: greater than 1 — false
   • D: any real number — false

Therefore, the correct option is $\boxed{\text{A}}.$