7. In $\triangle \mathrm{ABC}$, $\operatorname{tg} \angle \mathrm{CAB}=\frac{22}{7}$, a line is drawn from $\mathrm{A}$ to $\mathrm{BC}$ dividing $BC$ into segments of lengths 3 and 17. What is the area of $\triangle \mathrm{ABC}$?
7. (110)

Let $x$ be the length of the altitude, intersecting $B C$ at $D$. Then, from $\angle B A C=\angle B A D+\angle D A C$, we get $\operatorname{arctg} \frac{22}{7}=\operatorname{arctg} \frac{3}{x}+\operatorname{arctg} \frac{17}{x}$.
Taking the tangent of both sides of (1) and using the $\operatorname{tg}(\alpha+\beta)$ formula, we obtain
$$
\frac{22}{7}=\frac{\frac{3}{x}+\frac{17}{x}}{1-\left(\frac{3}{x}\right)\left(\frac{17}{x}\right)}
$$

Simplifying, we get
$$
11 x^{2}-10 x-51=(x-11)(11 x+51)=0 \text {. }
$$

Since $x>0$, we know $11 x+51>0$, thus $x=11$.
Therefore, $S_{\triangle A B C}=\frac{1}{2}(3+17) \times 11=110$.