If $a^2+b^2+c^2=ab+bc+ca$, then which one is correct?

Given the algebraic equation:
$$a^2+b^2+c^2=ab+bc+ca$$

Rearranging all terms to one side:

$$a^2+b^2+c^2-ab-bc-ca=0$$

Multiplying the entire equation by $2$:

$$2a^2+2b^2+2c^2-2ab-2bc-2ca=0$$

Regrouping the terms into perfect squares:

$$(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0$$

$$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$$

Since the square of any real number is always non-negative ($\ge 0$), the sum of three squared terms can only be equal to zero if each individual term is independently zero:

1. $(a-b{)}^2=0⟹a=b$
2. $(b-c{)}^2=0⟹b=c$
3. $(c-a{)}^2=0⟹c=a$

Combining these results, we get:

$$a=b=c$$