In triangle $ABC$, $\tan \angle CAB = \frac{22}{7}$, and the altitude from $A$ divides $\overline{BC}$ into segments of length $3$ and $17$. What is the area of triangle $ABC$?
Let $h = AD.$

[asy]
unitsize(0.3 cm);

pair A, B, C, D;

A = (3,11);
B = (0,0);
D = (3,0);
C = (20,0);

draw(A--B--C--cycle);
draw(A--D);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$3$", (B + D)/2, S);
label("$17$", (C + D)/2, S);
label("$h$", (A + D)/2, E);
[/asy]

Then $\tan \angle BAD = \frac{3}{h}$ and $\tan \angle CAD = \frac{17}{h},$ so
\begin{align*}
\tan A &= \tan (\angle BAD + \angle CAD) \\
&= \frac{\tan \angle BAD + \tan \angle CAD}{1 - \tan \angle BAD \cdot \tan \angle CAD} \\
&= \frac{\frac{3}{h} + \frac{17}{h}}{1 - \frac{3}{h} \cdot \frac{17}{h}} \\
&= \frac{20h}{h^2 - 51}.
\end{align*}Thus, $\frac{20h}{h^2 - 51} = \frac{22}{7}.$  This simplifies to
\[11h^2 - 70h - 561 = 0,\]which factors as $(h - 11)(11h + 51) = 0.$  Hence, $h = 11,$ and the area of triangle $ABC$ is $\frac{1}{2} \cdot 20 \cdot 11 = \boxed{110}.$