3. Given that the area of parallelogram $A B C D$ is $240, E, H$ are the midpoints of sides $A D, A B$ respectively, point $G$ is on side $B C$ such that $B G=2 G C$, point $F$ is on side $C D$ such that $D F=3 F C$, point $K$ is on side $A C$ such that the area of $\triangle E K F$ is 33. Find the area of $\triangle H K G$.
3. 32 .

Given that $AC$ is a diagonal of parallelogram $ABCD$,
$S_{\triangle ACD}=240 \times \frac{1}{2}=120$.
Assume $\frac{AK}{KC}=\frac{1}{b}$.
By the Angle Bisector Theorem,
$$
\begin{array}{l}
S_{\triangle DEF}=120 \times \frac{1}{2} \times \frac{3}{4}=45, \\
S_{\triangle AKE}=120 \times \frac{1}{2} \times \frac{1}{1+b}=\frac{60}{1+b}, \\
S_{\triangle CFK}=120 \times \frac{b}{1+b} \times \frac{1}{4}=\frac{30b}{1+b} .
\end{array}
$$

Then $33+45+\frac{60}{1+b}+\frac{30b}{1+b}=120$
$$
\begin{array}{l}
\Rightarrow 60+30b=42(1+b) \\
\Rightarrow b=\frac{3}{2} .
\end{array}
$$

Also, $S_{\triangle ACB}=240 \times \frac{1}{2}=120$, thus, by the Angle Bisector Theorem,
$$
\begin{array}{l}
S_{\triangle BGH}=120 \times \frac{1}{2} \times \frac{2}{3}=40, \\
S_{\triangle AKH}=120 \times \frac{1}{2} \times \frac{1}{1+b}=\frac{60}{1+b}=24, \\
S_{\triangle CGK}=120 \times \frac{b}{1+b} \times \frac{1}{3}=\frac{40b}{1+b}=24 .
\end{array}
$$

Therefore, $S_{\triangle HKG}=120-40-24-24=32$.