6. (3 points) Given a triangle $ABC$ with sides $AB=21, BC=20, AC=13$. On side $BC$, point $N$ is marked; on side $AB$, point $L$ is marked; on side $AC$, point $K$ is marked. It is known that $AK=1, CN=12, AL=\frac{13}{21}$. A line parallel to $KL$ is drawn through point $N$, intersecting side $AB$ at point $M$. Find the area of quadrilateral $KLMN$.
Answer: $\frac{14236}{325}=43 \frac{161}{325}$