Solution:

The answer is (A). Let $I$ be the incenter of triangle $ABC$; then triangles $AIB$ and $XIY$ are similar because of the parallelism between the lines $AB$ and $XY$. Let $h$ be the height of the trapezoid $ABXY$, that is, the height of triangle $ABC$ relative to the base $AB$; the height of triangle $AIB$ relative to $AB$ is the radius of the circle inscribed in $ABC$, whose length is $S_{ABC} / p_{ABC} = S_{ABC} / 34$. On the other hand, $h = 2 S_{ABC} / 17$. We have, thanks to the similarity mentioned above, $XY : AB = XY : 17 = \left(2 S_{ABC} / 17 - S_{ABC} / 34\right) : S_{ABC} / 34$; that is, $XY = 3 \cdot 17 = 51$. The area of the trapezoid is therefore $\frac{1}{2}(17+51) h = 34 h$.

The height $h$ can be computed as $\frac{2}{AB} \sqrt{p(p-AB)(p-BC)(p-AC)}$, using Heron's formula, where $p$ is the semiperimeter of $ABC$; substituting the side lengths gives $h = 24$, and thus $S_{ABXY} = 816$. Alternatively, it is possible to compute $h$ using the Pythagorean Theorem: letting $x$ and $y$ be the projections of the height onto $AB$, we know that $x+y=17$, $x^{2}+h^{2}=25^{2}$ and $y^{2}+h^{2}=26^{2}$, from which $y^{2}-x^{2}=(x+y)(y-x)=17(x-y)=51$. Solving the system for $x$ and $y$ we obtain, for instance, $y=10$, and hence $h=\sqrt{26^{2}-10^{2}}=24$.