For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax + by +…

$\left(\text{a}-\text{b}\right)\text{x}+\left(\text{b}-\text{a}\right)\text{y}=0$
$\Rightarrow \text{x}=\text{y}$
$\Rightarrow \text{Point of intersection}\left(\frac{-\text{c}}{\text{a}+\text{b}}\text{,}\frac{-\text{c}}{\text{a}+\text{b}}\right)$
Now using distance formula between two points $d=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

$\Rightarrow \sqrt{{\left(1+\frac{\text{c}}{\text{a}+\text{b}}\right)}^2+{\left(1+\frac{\text{c}}{\text{a}+\text{b}}\right)}^2}<2\sqrt{2}$
$\Rightarrow \sqrt{2}\left(\left|\frac{\text{a}+\text{b}+\text{c}}{\text{a}+\text{b}}\right|\right)<2\sqrt{2}$

$\Rightarrow \left(\frac{\text{a}+\text{b}+\text{c}}{\text{a}+\text{b}}\right)<2$

$\text{⇒a}+\text{b}+\text{c}<2\left(\text{a}+\text{b}\right)$
$\Rightarrow \text{a}+\text{b}-\text{c}>0$