jmc

We begin by drawing a diagram and labeling the vertices $A$, $B$, and $C$ as shown:



We drop a perpendicular from $C$ to $\overline{AB}$ and label the intersection point $X$. $X$ divides $AB$ into two segments $AX$ and $XB$ that sum to 17, so let $AX=x$ and $XB=17-x$. Let the height of the triangle, $CX$, have length $h$.



Now we have two right triangles, so we can use the Pythagorean theorem on both triangles to write two equations in terms of $x$ and $h$. From $△AXC$, we have $$x^2+h^2=13^2,$$ and from $△CXB$, we have $$(17-x{)}^2+h^2=(12\sqrt{2}{)}^2.$$ Expanding the second equation gives $289-34x+x^2+h^2=144\cdot 2=288$; substituting the first equation into the second gives $$289-34x+13^2=288.$$ Simplifying and solving for $x$ yields $1+169=34x$, so $x=170/34=5$. Plugging this value into the first equation gives $$h^2=13^2-x^2=169-25=144,$$ so $h=\sqrt{144}=12$. Finally, we can compute the area of $△ABC$ to be $$\frac{1}{2}(AB)(h)=\frac{1}{2}(17)(12)=\boxed{102}.$$