3. From point $M$, lying inside triangle $A B C$, perpendiculars are drawn to the sides $B C, A C, A B$, with lengths $k, l$, and $m$ respectively. Find the area of triangle $A B C$, if $\angle C A B=\alpha$ and $\angle A B C=\beta$. If the answer is not an integer, round it to the nearest integer.

$$
\alpha=\frac{\pi}{6}, \beta=\frac{\pi}{4}, k=3, l=2, m=4
$$
Answer: 67.

Solution. Denoting the sides of the triangle by $a, b, c$, using the sine theorem we get $S=\frac{k a+l b+m c}{2}=R(k \sin \alpha+l \sin \beta+m \sin \gamma)$.

Since, in addition, $S=2 R^{2} \sin \alpha \sin \beta \sin \gamma$, we can express $R=\frac{k \sin \alpha+l \sin \beta+m \sin \gamma}{2 \sin \alpha \sin \beta \sin \gamma}$.

Therefore, $S=\frac{(k \sin \alpha+l \sin \beta+m \sin \gamma)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}$.

After substituting the numerical values of $k, l, m, \alpha, \beta, \gamma$ we get

$$
S=\frac{(3 \sin \alpha+2 \sin \beta+4 \sin \gamma)^{2}}{2 \sin \alpha \sin \beta \sin \gamma}=\frac{(3+4 \sqrt{2}+2 \sqrt{6})^{2}}{\sqrt{3}+1} \approx 67
$$