Step 1: Identify the key geometric relationship.
The points are $A(2, 0)$, $B(0, 2)$, and $P(1, 1)$. First check whether $P$ has any special position relative to $A$ and $B$.
Step 2: Show that $P$ is the midpoint of $AB$.
The midpoint of $A(2,0)$ and $B(0,2)$ is $\left(\frac{2+0}{2}, \frac{0+2}{2}\right) = (1, 1) = P$. So $P$ is exactly the midpoint of segment $AB$. This is the crucial observation.
Step 3: Recall the algebraic distance concept.
For a line $L: ax + by + c = 0$, the algebraic (signed) distance of a point $(x_0, y_0)$ from $L$ is $\frac{ax_0 + by_0 + c}{\sqrt{a^2+b^2}}$. Points on the same side of $L$ have the same sign; points on opposite sides have opposite signs.
Step 4: Analyse the distances of $A$ and $B$ from any line through $P$.
Let the line through $P$ be $ax + by + c = 0$. Since $P = (1,1)$ lies on this line, $a(1) + b(1) + c = 0$, i.e. $a + b + c = 0$. The algebraic distance of $A(2,0)$ is $\frac{2a + c}{\sqrt{a^2+b^2}}$ and of $B(0,2)$ is $\frac{2b + c}{\sqrt{a^2+b^2}}$.
Step 5: Compute the sum $d$.
\[ d = \frac{2a + c + 2b + c}{\sqrt{a^2+b^2}} = \frac{2(a+b) + 2c}{\sqrt{a^2+b^2}} = \frac{2(a+b+c)}{\sqrt{a^2+b^2}} \] Since $a + b + c = 0$ (because the line passes through $P$), we get $d = 0$.
Step 6: Note this holds for every line through $P$.
The result $d = 0$ follows purely from $a + b + c = 0$, which is true for every line passing through $P(1,1)$. So $d = 0$ for all such lines.
Step 7: State the final answer.
\[ \boxed{d = 0 \text{ for all lines through } P} \]