If $$\frac{a^2+b^2}{c^2}=\frac{b^2+c^2}{a^2}=\frac{c^2+a^2}{b^2}=\frac{1}{k}, (k\neq0)$$ then k=?

Given : $$\frac{a^2+b^2}{c^2}=\frac{b^2+c^2}{a^2}=\frac{c^2+a^2}{b^2}=\frac{1}{k}, (k\neq0)$$

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

Similarly, $$b^2+c^2=\frac{a^2}{k}$$ ------------(ii)

and $$c^2+a^2=\frac{b^2}{k}$$ ----------(iii)

Adding equations (i),(ii) and (iii)

=> $$2(a^2+b^2+c^2)=\frac{a^2}{k}+\frac{b^2}{k}+\frac{c^2}{k}=\frac{a^2+b^2+c^2}{k}$$

=> $$2=\frac{1}{k}$$

=> $$k=\frac{1}{2}$$

=> Ans - (D)