**Solution:**

To calculate the normality of the resulting solution, we need to determine the number of moles of the acid and base involved in the reaction. Then we can use the stoichiometry of the reaction to find the normality.

**Step 1: Calculate the number of moles of H3PO4**

Given:
Volume of H3PO4 solution = 500 mL = 0.5 L
Molarity of H3PO4 solution = 0.1 M

Using the formula:

Number of moles = Volume (L) x Molarity

Number of moles of H3PO4 = 0.5 L x 0.1 M = 0.05 moles

**Step 2: Calculate the number of moles of Ca(OH)2**

Given:
Volume of Ca(OH)2 solution = 250 mL = 0.25 L
Molarity of Ca(OH)2 solution = 0.05 M

Using the formula:

Number of moles = Volume (L) x Molarity

Number of moles of Ca(OH)2 = 0.25 L x 0.05 M = 0.0125 moles

**Step 3: Determine the stoichiometry of the reaction**

The balanced chemical equation for the reaction between H3PO4 and Ca(OH)2 is:

2 H3PO4 + 3 Ca(OH)2 → Ca3(PO4)2 + 6 H2O

From the balanced equation, we can see that 2 moles of H3PO4 react with 3 moles of Ca(OH)2.

**Step 4: Calculate the normality of the resulting solution**

The normality is defined as the number of equivalent moles of solute per liter of solution. In this case, since the reaction involves a 1:1 ratio of H3PO4 and Ca(OH)2, we can say that the number of equivalent moles is the same as the number of moles of H3PO4.

Normality = Number of moles / Volume (L)

Normality = 0.05 moles / 0.5 L = 0.1 N

Therefore, the normality of the resulting solution is 0.1 N.

In summary, to calculate the normality of the resulting solution, we determined the number of moles of H3PO4 and Ca(OH)2 using their given molarities and volumes. Then, using the stoichiometry of the reaction, we found that the number of equivalent moles is the same as the number of moles of H3PO4. Finally, we divided the number of moles by the volume of the solution to find the normality, which is 0.1 N.