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(D) Given that $\frac{a \sqrt{2}+b}{b \sqrt{2}+c} = k$,where $k \in \mathbb{Q}$.Then $a \sqrt{2} + b = k(b \sqrt{2} + c) = kb \sqrt{2} + kc$.Equating

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$\frac{a^2+b^2+c^2}{a+b+c}$

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(D) Given that $\frac{a \sqrt{2}+b}{b \sqrt{2}+c} = k$,where $k \in \mathbb{Q}$.Then $a \sqrt{2} + b = k(b \sqrt{2} + c) = kb \sqrt{2} + kc$.Equating the rational and irrational parts,we get $a = kb$ and $b = kc$.Thus,$\frac{a}{b} = \frac{b}{c} = k$,which implies $a, b, c$ are in geometric progression.Let $b = ar$ and $c = ar^2$ for some rational $r = \frac{b}{a}$. Since $a, b, c$ are integers,$r$ must be such that $b$ and $c$ are integers.Substituting $b = ar$ and $c = ar^2$ into the expression $\frac{a^2+b^2+c^2}{a+b+c}$:$= \frac{a^2 + a^2r^2 + a^2r^4}{a + ar + ar^2} = \frac{a^2(1+r^2+r^4)}{a(1+r+r^2)}$.Since $1+r^2+r^4 = (1+r^2)^2 - r^2 = (1+r^2-r)(1+r^2+r)$,the expression simplifies to $a(1-r+r^2)$.Given $k = \frac{a}{b} = \frac{1}{r}$,so $r = \frac{b}{a}$.Substituting $r = \frac{b}{a}$ into $a(1 - \frac{b}{a} + \frac{b^2}{a^2}) = a - b + \frac{b^2}{a}$.For this to be an integer for all such $a, b, c$,we observe that $b^2 = ac$,so $\frac{b^2}{a} = c$.Thus,$a - b + c$ is always an integer.

Let $a, b, c$ be positive integers such that $\frac{a \sqrt{2}+b}{b \sqrt{2}+c}$ is a rational number,then which of the following is always an integer?

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