Let $AB$ be of length $6$. The $VM=4,$ $VN=3,$ $VP=2.$ Let $O$ be the intersection of $MP$ and $QN.$ Then $O$ lies on the altitude of the pyramid, which also bisects the right angles $\angle MVP$ and $\angle QVN.$ Let points $Q'$ and $N'$ lie on segments $VA$ and $VC$, respectively, such that $VQ'=VQ$ and $VN'=VN$. Then $Q'N'$ and $MP$ also intersect at $H.$ We have now simplified a 3D problem into a 2D problem and can be graphed like so: [asy][asy]defaultpen(fontsize(10pt)); pair V,Q,M,A,H,P,N,C; V=(0,0);Q=(0,12/5);M=(0,4);A=(0,6);H=(4/3,4/3);P=(2,0);N=(3,0);C=(6,0);dot("$V(0,0)$",V,SW);dot("$A(0,6)$",A,W);dot("$M(4,0)$",M,W);dot("$Q'$",Q,W);draw(A--V); dot("$P(2,0)$",P,S);dot("$N'(3,0)$",N,S);dot("$C(6,0)$",C,S);draw(C--V--H);draw(Q--N);draw(M--P);dot("$O$",H,NE);[/asy][/asy] Now we can do some light coordinate bashing to find that $O=\left(\tfrac43,\tfrac43\right),$ and then from there $Q'=\left(0,\tfrac{12}5\right).$ So $VQ:QD=VQ':Q'A=\frac{12}5:6-\frac{12}5=\fbox{2:3}.$