If $$a^{2}+b^{2}+c^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=6$$, then what is the value of $$(a^{2}+b^{2}+c^{2})$$ ?

Given : $$a^{2}+b^{2}+c^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=6$$

=> $$(a^2+\frac{1}{a^2}-2)+(b^2+\frac{1}{b^2}-2)+(c^2+\frac{1}{c^2}-2)=0$$

=> $$(a-\frac{1}{a})^2+(b-\frac{1}{b})^2+(c-\frac{1}{c})^2=0$$

$$\because$$ Sum of three positive terms is zero, hence each term is equal to 0.

=> $$(a-\frac{1}{a})=$$ $$(b-\frac{1}{b})=$$ $$(c-\frac{1}{c})=0$$

=> $$\frac{a^2-1}{a}=0$$

=> $$a^2=1$$

Similarly, $$b^2=c^2=1$$

$$\therefore$$ $$(a^{2}+b^{2}+c^{2})=1+1+1=3$$ 

=> Ans - (A)