1. Since $3,a,b,c$ are in A.P., let the common difference be $d$.

   Then $a=3+d,\quad b=3+2d,\quad c=3+3d.$

2. Now $3,\ a-1,\ b+1,\ c+9$ are in G.P.

   Substitute the A.P. values: $a-1=(3+d)-1=2+d,$ $b+1=(3+2d)+1=4+2d,$ $c+9=(3+3d)+9=12+3d.$

   So the four numbers $3,\ 2+d,\ 4+2d,\ 12+3d$ are in G.P.

3. For consecutive terms of a G.P., the square of the middle term equals the product of its neighbors.

   Thus, $(2+d)^2 = 3(4+2d).$

   Expanding: $d^2+4d+4=12+6d$ $d^2-2d-8=0$ $(d-4)(d+2)=0.$

   Hence, $d=4 \quad \text{or} \quad d=-2.$

4. Use the second G.P. condition: $(4+2d)^2=(2+d)(12+3d).$

   Check $d=4$: $4+2d=12,\quad 2+d=6,\quad 12+3d=24$ $12^2=144,\qquad 6\cdot 24=144$ so $d=4$ works.

   Check $d=-2$: $2+d=0,\quad 4+2d=0,\quad 12+3d=6$ This gives terms $3,0,0,6$, which cannot form a G.P. because the common ratio is not defined consistently.

   Therefore, only $d=4$ is valid.

5. Now find $a,b,c$: $a=3+4=7,\quad b=3+8=11,\quad c=3+12=15.$

6. Their arithmetic mean is $\frac{a+b+c}{3}=\frac{7+11+15}{3}=\frac{33}{3}=11.$

Therefore, the required arithmetic mean is $\boxed{11}.$