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(B) Let the equation of the variable line be $ax + by + c = 0$.The perpendicular distance from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is g

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pass through the fixed point $(1, 1)$.

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(B) Let the equation of the variable line be $ax + by + c = 0$.The perpendicular distance from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is given by $d = \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}}$.Given that the algebraic sum of the perpendicular distances from $A(2, 0)$,$B(0, 2)$,and $C(1, 1)$ is zero:$\frac{2a + c}{\sqrt{a^2 + b^2}} + \frac{2b + c}{\sqrt{a^2 + b^2}} + \frac{a + b + c}{\sqrt{a^2 + b^2}} = 0$Multiplying by $\sqrt{a^2 + b^2}$ (assuming $\sqrt{a^2 + b^2} 
eq 0$):$(2a + c) + (2b + c) + (a + b + c) = 0$Combining like terms:$3a + 3b + 3c = 0$Dividing by $3$:$a + b + c = 0$This implies $c = -(a + b)$. Substituting this into the line equation $ax + by + c = 0$:$ax + by - (a + b) = 0$$a(x - 1) + b(y - 1) = 0$This equation holds for all $a$ and $b$ if $x - 1 = 0$ and $y - 1 = 0$,which means $x = 1$ and $y = 1$.Thus,all such lines pass through the fixed point $(1, 1)$.

Let the algebraic sum of the perpendicular distances from the points $A(2, 0)$,$B(0, 2)$,and $C(1, 1)$ to a variable line be zero. Then,all such lines:

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