Question:

Step 1: Find the semi-perimeter $s$ using the given conditions. We are given that $s - a = 3$, $s - b = 5$, and $s - c = 7$. Adding these three equations: $$(s-a) + (s-b) + (s-c) = 3 + 5 + 7$$ $$3s - (a+b+c) = 15$$ Since $a + b + c = 2s$, we have: $$3s - 2s = 15$$ $$s = 15$$ Step 2: Determine the side lengths of the triangle. Using $s = 15$ and the given conditions: $$a = s - (s-a) = 15 - 3 = 12$$ $$b = s - (s-b) = 15 - 5 = 10$$ $$c = s - (s-c) = 15 - 7 = 8$$ Step 3: Calculate the area of triangle $ABC$ using Heron's formula. $$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{15 \cdot 3 \cdot 5 \cdot 7}$$ $$\Delta = \sqrt{1575} = \sqrt{225 \cdot 7} = 15\sqrt{7}$$ Step 4: Find the inradius of the triangle. The inradius is given by: $$r = \frac{\Delta}{s} = \frac{15\sqrt{7}}{15} = \sqrt{7}$$ Step 5: Identify the key properties of quadrilateral $QCRI$. The incircle touches side $BC$ at $Q$ and side $CA$ at $R$. By the property of tangents from an external point to a circle: $$CQ = CR = s - c = 7$$ Since $I$ is the incenter, the radii $IQ$ and $IR$ are perpendicular to the sides $BC$ and $CA$ respectively: $$IQ \perp BC \text{ and } IR \perp CA$$ Both radii equal the inradius: $$IQ = IR = r = \sqrt{7}$$ Step 6: Calculate the area of quadrilateral $QCRI$. The quadrilateral $QCRI$ can be divided into two triangles: $\triangle QCI$ and $\triangle RCI$. For $\triangle QCI$: The base is $CQ = 7$ and the height (perpendicular distance from $I$ to $BC$) is $IQ = \sqrt{7}$. $$\text{Area of } \triangle QCI = \frac{1}{2} \cdot CQ \cdot IQ = \frac{1}{2} \cdot 7 \cdot \sqrt{7}$$ For $\triangle RCI$: The base is $CR = 7$ and the height (perpendicular distance from $I$ to $CA$) is $IR = \sqrt{7}$. $$\text{Area of } \triangle RCI = \frac{1}{2} \cdot CR \cdot IR = \frac{1}{2} \cdot 7 \cdot \sqrt{7}$$ Therefore, the total area of quadrilateral $QCRI$ is: $$\text{Area of } QCRI = \frac{1}{2} \cdot 7 \cdot \sqrt{7} + \frac{1}{2} \cdot 7 \cdot \sqrt{7} = \frac{1}{2} \cdot \sqrt{7}(7 + 7) = \frac{1}{2} \cdot \sqrt{7} \cdot 14 = 7\sqrt{7}$$ **Final Answer:** The area of quadrilateral $QCRI$ is $\boxed{7\sqrt{7}}$, which corresponds to **Option 4**.