Let P is the point of intersection of line $$a x+b y +c=0$$ and $$b x+a y-c=0$$

$$\mathrm{P}=\left(\frac{-c}{a+b}, \frac{-c}{a+b}\right)$$

Given, the distance between $$(1,1)$$ and $$P$$ is less than $$2 \sqrt{2}$$

$$\begin{aligned} & \therefore \sqrt{\left(1+\frac{c}{a+b}\right)^2+\left(1+\frac{c}{a+b}\right)^2}<2 \sqrt{2} \\ & \Rightarrow \sqrt{2}\left|\frac{a+b+c}{a+b}\right|<2 \sqrt{2} \\ & \Rightarrow\left|\frac{a+b+c}{a+b}\right|<2 \\ & \therefore a>b>c>0 \\ & \therefore \frac{a+b+c}{a+b}<2 \\ & \Rightarrow a+b+c<2 a+2 b \\ & \Rightarrow a+b-c>0 \end{aligned}$$

Hence, option (A) correct

$$\therefore$$ Given $$a > b > c > 0$$

$$\therefore a-b$$ is positive and c is also positive

$$\Rightarrow a-b+c>0$$

Hence, option (C) is also true.

Hints :

Given, $$a > b > c > 0$$

So, $$a-b, a-c, b$$ and $$c$$ all are positive

$$\therefore \quad a-b+c>0, a-c+b>0$$