Will do the job in this case, but is not massively robust. Would need additional code to enforce the number of elements in the vector being reshaped if there is a chance it will not be a multiple of (six in this case)
vec=[0 0 0 3 5 7 2 1 4 0 0 0 4 7 8 5 2 4];
%Remove zeros
vec(vec==0)=[];
index=1;
for iter=1:6:numel(vec)
%Reshape remaining vector into a 3x2 and store
mat(:,:,index)=reshape(vec(iter:iter+5),[],2);
index=index+1;
end
mat(:,:,1)
mat(:,:,2)
