To solve the problem, we start by understanding the relationships between the triangles and their respective areas using the properties of medians and midpoints.
Triangle \( JKL \) has points \( M, N, \) and \( O \) as the midpoints of sides \( KL, LJ, \) and \( JK \), respectively. Therefore, triangle \( MNO \) is formed by these midpoints.
According to the midpoint theorem, the area of triangle \( MNO \) will be one quarter of the area of triangle \( JKL \). If we denote the area of triangle \( JKL \) as \( A \), we can express the area of triangle \( MNO \) as:
\[ \text{Area of } MNO = \frac{1}{4} A \]
Next, we have points \( P, Q, \) and \( R \) as midpoints of segments \( NO, OM, \) and \( MN \), respectively. Triangle \( PQR \) is formed by these new midpoints.
Again applying the midpoint theorem, we can determine the relationship between the area of triangle \( PQR \) and the area of triangle \( MNO \). Specifically, the area of triangle \( PQR \) will be one quarter of the area of triangle \( MNO \):
\[ \text{Area of } PQR = \frac{1}{4} \cdot \text{Area of } MNO \]
Substituting the earlier expression for the area of \( MNO \) gives:
\[ \text{Area of } PQR = \frac{1}{4} \left( \frac{1}{4} A \right) = \frac{1}{16} A \]
We know from the problem statement that the area of triangle \( PQR \) is 12:
\[ \frac{1}{16} A = 12 \]
To find the area \( A \), we solve for \( A \) by multiplying both sides of the equation by 16:
\[ A = 12 \times 16 = 192 \]
The area of triangle \( JKL \) is \( A = 192 \). Now, we are tasked with finding the area of triangle \( JQR \).
Since \( JQR \) shares point \( J \) with triangle \( JKL \) and points \( Q \) and \( R \) are midpoints of sides of a smaller triangle \( MNO \), the relationship holds that triangle \( JQR \) is half of triangle \( JKL \).
To express this, we will use the area ratio established by the midpoints:
The area of triangle \( JQR \) can be determined as follows:
\[ \text{Area of } JQR = \frac{1}{2} \cdot \text{Area of } MNO = \frac{1}{2} \cdot \frac{1}{4} A = \frac{1}{8} A \]
Substituting the area \( A = 192 \):
\[ \text{Area of } JQR = \frac{1}{8} \cdot 192 = 24 \]
Thus, the area of triangle \( JQR \) is
\[ \boxed{24} \]