For real a, b, c if $$a^2 + b^2 + c^2 = ab + bc + ca$$, the value of $$\frac{(a+b)}{c}$$

Given,    $$a^2+b^2+c^2=ab+bc+ca$$

Multiplying both sides by "2", it becomes

$$2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)$$

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

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

$$\left(a^2+b^2-2ab\right)+\left(b^2+c^2-2ca\right)+\left(c^2+a^2-2ca\right)=0$$

$$\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0$$

Sum of squares is zero so each term should be zero

$$=$$>    $$\left(a-b\right)^2=0$$, $$\left(b-c\right)^2=0$$, $$\left(c-a\right)^2=0$$

$$=$$>      $$a-b=0$$,       $$b-c=0$$,       $$c-a=0,$$

$$=$$>         $$a=b$$,               $$b=c$$,                $$c=a$$

$$=$$>         $$a=b=c$$

Therefore     $$\ \frac{\ a+b}{c}=\ \frac{\ a+a}{a}=\ \frac{\ 2a}{a}=2$$