Question 1210493
This is a geometry problem involving the **medial triangle** and the relationship between the areas of nested triangles formed by connecting midpoints.

The core principle used here is the **Medial Triangle Theorem** as it relates to area.

## 1. Defining the Triangles

Let $T_0$ be the original triangle, $T_1$ be its medial triangle, and $T_2$ be the medial triangle of $T_1$, and so on.

1.  **Original Triangle:** $\triangle JKL$
2.  **First Medial Triangle ($T_1$):** $\triangle MNO$. Its vertices ($M, N, O$) are the midpoints of the sides of $\triangle JKL$.
3.  **Second Medial Triangle ($T_2$):** $\triangle PQR$. Its vertices ($P, Q, R$) are the midpoints of the sides of $\triangle MNO$. ($\overline{NO}, \overline{OM}, \overline{MN}$).
    * Given: $\text{Area}(\triangle PQR) = 12$.
4.  **Third Medial Triangle ($T_3$):** $\triangle XYZ$. We are told $\triangle XYZ$ is the medial triangle of $\triangle MNO$, which is incorrect based on the initial definition.
    * **Correction:** Based on the standard progression, $T_1 = \triangle MNO$. The next medial triangle is $\triangle PQR$. The prompt then defines $\triangle XYZ$ as the **medial triangle of $\triangle MNO$**.
    * **Reconciliation:** This means $\triangle XYZ$ is the same as $\triangle PQR$ ($T_2$).
    * $\text{Area}(\triangle XYZ) = \text{Area}(\triangle PQR)$.

## 2. Medial Triangle Area Property

The area of the **medial triangle** (the triangle formed by connecting the midpoints of the sides) is always **one-fourth ($\frac{1}{4}$)** the area of the original triangle. 

$$\text{Area}(T_{k+1}) = \frac{1}{4} \cdot \text{Area}(T_k)$$

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## 3. Determining the Area of $\triangle XYZ$

We are given $\triangle PQR$ (which is $T_2$) and asked for $\triangle XYZ$, which is defined as the medial triangle of $\triangle MNO$ (which is $T_1$).

Since $\triangle PQR$ is also defined as the medial triangle of $\triangle MNO$ (its vertices $P, Q, R$ are the midpoints of $NO, OM, MN$), we must assume that **$\triangle XYZ$ is the same triangle as $\triangle PQR$**.

$$\triangle XYZ = \triangle PQR$$

Therefore, the area of $\triangle XYZ$ is the same as the area of $\triangle PQR$.

$$\text{Area}(\triangle XYZ) = \text{Area}(\triangle PQR)$$
$$\text{Area}(\triangle XYZ) = \mathbf{12}$$

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### Supplementary Calculation (If $\triangle XYZ$ was the Medial Triangle of $\triangle PQR$)

If the question had intended the sequence $T_0 \to T_1 \to T_2 \to T_3$, where $\triangle XYZ$ was the medial triangle of $\triangle PQR$:

$$\text{Area}(\triangle XYZ) = \frac{1}{4} \cdot \text{Area}(\triangle PQR) = \frac{1}{4} \cdot 12 = 3$$