In $\triangle ABC$, $AB=15$, $AC=9$, and $BC=12$. Points $M$ and $N$ lie on $AC$ and $BC$, respectively, with $CM=CN=3$. Points $J$ and $K$ are on $AB$ so that $MJ$ and $NK$ are perpendicular to $AB$. What is the area of pentagon $CMJKN$?

$\triangle BKN\sim \triangle BCA$ with ratio $BN:BA=(12-3):15=9:15=3:5$, so $[BKN]:[BAC]=9:25$.
$\triangle AMJ\sim \triangle ABC$ with ratio $AM:AB=(9-3):15=6:15=2:5$, so $[AMJ]:[BAC]=4:25$.
$([BKN]+[AMJ]):[ABC]=13:25$, so $[CNKJM]:[ABC]=12:25$.
$[ABC]=\frac12(9)(12)=54$, so $[CNKJM]=\frac{12}{25}(54)=\boxed{\frac{648}{25}}$.