jmc

Since $△ABC$ is equilateral with side length 12 and $X$ and $Y$ are the midpoints of $CA$ and $CB$ respectively, we have $CX=CY=\frac{1}{2}(12)=6$. Since the height of the prism is 16 and $Z$ is the midpoint of $CD$ we have $CZ=\frac{1}{2}(16)=8$.

We have $\angle ACD=\angle BCD=90^{\circ }$ since faces $ACDE$ and $BCDF$ are rectangles. Thus, $△XCZ$ and $△YCZ$ are right-angled at $C$. By the Pythagorean Theorem, $$XZ=\sqrt{C{X}^2+C{Z}^2}=\sqrt{6^2+8^2}=\sqrt{100}=10$$and $$YZ=\sqrt{C{Y}^2+C{Z}^2}=\sqrt{6^2+8^2}=\sqrt{100}=10.$$Now we look at $△CXY$. We know that $CX=CY=6$ and that $\angle XCY=60^{\circ }$, because $△ABC$ is equilateral. Thus, $△CXY$ is isosceles with $\angle CXY=\angle CYX$. These angles must each be equal to $\frac{1}{2}(180^{\circ }-\angle XCY)=\frac{1}{2}(180^{\circ }-60^{\circ })=60^{\circ }$. Thus $△CXY$ is equilateral, so $XY=CX=CY=6$.

Finally, $XY=6$ and $XZ=YZ=10$. The perimeter is then $10+10+6=\boxed{26}$.