The answer is $\boxed{19}$. First note that the line can only cut two sides of the triangle, so at least one side remains intact. The diameter of the plate must clearly be at least as long as this side, so it is at least $19$. Now we'll show that it's indeed possible to do it with a plate with diameter $19$. Label the vertices of the cake as $ABC$ so that $AB=19, BC=20, CA=21$ Let the circle $\omega$ with diameter $AB$ cut $AC,BC$ at $X,Y$. Cut the triangle along the line $XY$. We claim that the $\triangle CXY$ fits in the remaining semicircle of $\omega$. Project $X$ onto $BC$ at $Z$. Move $\triangle CXY$ to $\triangle C'X'Y'$ so that $Z'$ aligns with the center of $\omega$ and $C'Y'$ aligns with $AB$. Clearly, $C',Y'$ lie in $\omega$, and after some calculations, we will have $XZ = \frac{80\sqrt{33}}{49} < \frac{19}{2}$, so $Z'$ lies in $\omega$ and we're done.