11. As shown in the figure, in $\triangle \mathrm{ABC}$, $\mathrm{M}$ is the midpoint of side $\mathrm{AB}$, $\mathrm{N}$ is a trisection point on side $\mathrm{AC}$, and $\mathrm{CM}$ intersects $\mathrm{BN}$ at point K. If the area of $\triangle B C K$ is 1, then the area of $\triangle A B C$ is $\qquad$ .
【Analysis】Connect $\mathrm{AK}$, by the燕tail model (Note: This should be "by the tail model" or "by the cevian theorem", but "燕tail model" is kept as a literal translation)
Since $\mathrm{AM}=\mathrm{MB}$, we have $S_{\triangle A K C}=S_{\triangle B K C}=1$
Since $\mathrm{AN}=2 \mathrm{NC}$, we have $S_{\triangle A K B}=2 S_{\triangle C K B}=2$
Therefore, $S_{\triangle A C B}=S_{\triangle A K B}+S_{\triangle B K C}+S_{\triangle C K A}=2+1+1=4$