Algebra or Geometry

This was a nice troll question. Let $b=ar$ , $c=ar^2$ , $q=p+d$ and $r=p+2d$ . Thus, the second statement says $a^{p+d}(ar)^{p+2d}(ar^2)^p = a^{3p+3d} r^{3p+2d} = 29.$

Note that $a^r b^p c^q = a^{p+2d} (ar)^{p} (ar^2)^{p+d} = a^{3p+3d} r^{3p+2d} = 29$ from above. Thus, the answer is $\boxed{29}$ .

$Let ~~a=a,.....b=a*x,.....c=a*x^2.......................p=p,..........q=p+d,..........r=p+2d.\\ So~~a^p*b^q*c^r=a^p*(a*x)^{p+d}*(a*x^2)^{p+2d}=a^{3p}*a^{3d}~*~x^{3p}*x^{5d}=6.\\ Also~~a^q*b^r*c^p=a^{p+d}*(a*x)^{p+2d}*(a*x^2)^{p}=\color{#20A900}{a^{3p}*a^{3d}~*~x^{3p}*x^{2d}=29}\\ We~want~~a^r*b^p*c^q=a^{p+2d}*(a*x)^{p}*(a*x^2)^{p+d}=\color{#20A900}{a^{3p}*a^{3d}~*~x^{3p}*x^{2d}=\Large \color{#D61F06}{29}.}$