jmc

Since $r$ is a root of $x^2+5x+7=0,$ $r^2+5r+7=0.$

We have that $$\begin{array}{rl}(r-1)(r+2)(r+6)(r+3)& =(r-1)(r+6)(r+2)(r+3)\\ & =(r^2+5r-6)(r^2+5r+6)\\ & =(-13)(-1)=\boxed{13}.\end{array}$$