Suppose $$a, b, c$$ are in A.P. and $$a^2, 2b^2, c^2$$ are in G.P. If $$a < b < c$$ and $$a + b + c = 1$$, then $$9(a^{2}+b^{2}+c^{2})$$ is equal to _____________.

Let (a, b, c) be in A.P. ⇒$$(b=\frac{a+c}{2}).$$
Given ($$a^2,2b^2,c^2$$) are in G.P.

For three terms in G.P., the middle squared equals the product of extremes:
$$(2b^2)^2=a^2c^2$$

Substitute ($$b=\frac{a+c}{2}$$):

$$2b^2=2\left(\frac{a+c}{2}\right)^2=\frac{(a+c)^2}{2}$$

$$\left(\frac{(a+c)^2}{2}\right)^2=a^2c^2$$

$$\frac{(a+c)^4}{4}=a^2c^2$$

$$(a+c)^4=4a^2c^2$$

Taking square root:
$$(a+c)^2=\pm2ac$$

The positive case gives only trivial solutions, so take:
$$(a+c)^2=-2ac$$

$$a^2+2ac+c^2=-2ac\Rightarrow a^2+4ac+c^2=0$$

Let (s = a+c). Then:
$$s^2+2ac=0\Rightarrow ac=-\frac{s^2}{2}$$

a + b + c = 1

$$s+\frac{s}{2}=\frac{3s}{2}=1\Rightarrow s=\frac{2}{3}$$

$$ac=-\frac{(2/3)^2}{2}=-\frac{2}{9}$$

$$a^2+c^2=s^2-2ac=\frac{4}{9}-2\left(-\frac{2}{9}\right)=\frac{8}{9}$$

$$b^2=\left(\frac{s}{2}\right)^2=\frac{1}{9}$$

$$a^2+b^2+c^2=\frac{8}{9}+\frac{1}{9}=1$$

$$9(a^2+b^2+c^2)=9$$