If $A(-3, 5)$, $B(7, 12)$, $C(5, 3)$ and $D$ are the four vertices of parallelogram $ABCD$, what are the coordinates of point $D$?
We know where points $A$, $B$, and $C$ are, and we know that the shape is a parallelogram, so we know that the differences in the x-values and y-values between $B$ and $A$ must be the same as the differences in x-values and y-values between $C$ and $D$.

The solution is $\boxed{(-5, -4)}$, which keeps the vertices in alphabetical order, as shown in the first figure. Two other parallelograms are shown in figures 2 and 3 (where we number them from left to right), but they are not solutions because the points aren't in the correct order. They would be parallelograms $ACBD$ and $ABDC$, respectively. [asy]
import olympiad; import geometry; size(250); defaultpen(linewidth(0.8));
picture a,b,c;
xaxis(a,YZero(),-6,6,Ticks(beginlabel=false,Step=20,step=2));
yaxis(a,XZero(),-5,13,Ticks(beginlabel=false,Step=20,step=2));
xaxis(b,YZero(),-6,6,Ticks(beginlabel=false,Step=20,step=2));
yaxis(b,XZero(),-5,13,Ticks(beginlabel=false,Step=20,step=2));
xaxis(c,YZero(),-6,17,Ticks(beginlabel=false,Step=10,step=2));
yaxis(c,XZero(),-5,15,Ticks(beginlabel=false,Step=10,step=2));
pair A = (-3,5),B=(7,12),C=(5,3);
draw(a,A--B--C--(-5,-4)--cycle);
label(a,"$A$",A,NW); label(a,"$B$",B,NE); label(a,"$C$",C,SE); label(a,"$D$",(-5,-4),SW);
draw(b,A--C--B--(-1,14)--cycle);
label(b,"$A$",A,SW); label(b,"$B$",B,NE); label(b,"$C$",C,SE); label(b,"$D$",(-1,14),NW);
draw(c,A--C--(15,10)--B--cycle);
label(c,"$A$",A,W); label(c,"$B$",B,N); label(c,"$C$",C,S); label(c,"$D$",(15,10),E);
add(currentpicture,a);
add(currentpicture, shift(20,0)*b);
add(currentpicture, shift(40,0)*c);
[/asy]