A.P-G.P

Since $a_1,a_2,a_3$ is an arithmetic progression with $a_1+a_2+a_3 = 15$ , we can conclude $a_1=5-x, a_2=5, a_3=5+x$ for some real number $x$ (the common difference). Similarly, since $b_1,b_2,b_3$ is a geometric progression with $b_1b_2b_3=27$ , we can conclude $b_1=\frac{3}{y}, b_2=3, b_3= 3y$ for some real number $y$ (the common ratio). Finally, since $a_1+b_1, a_2+b_2, a_3+b_3$ is a geometric progression of positive integers, we can conclude $\left(5-x+\frac{3}{y}\right)\left(5+x+3y\right) = (a_1+b_1)(a_3+b_3) = (a_2+b_2)^2 = (5+3)^2 = 2^6$ and therefore $5-x+\frac{3}{y} = 2^k \quad , \quad 5+x+3y = 2^{6-k} \qquad \text{for some } k \in \{0,1,2,3,4,5,6\}.$

By eliminating $y$ from this system of two equations, we end up with the quadratic equation $x^2 + \left(2^k-2^{6-k}\right)x + 5\left(2^k+2^{6-k}\right) - 80 = 0$ and by the quadratic formula, $\begin{aligned}x &= \frac{1}{2}\left(-\left((2^k-2^{6-k}\right)\pm\sqrt{\left(2^k-2^{6-k}\right)^2 - 4\left(5\left(2^k+2^{6-k}\right)-80\right)}\right) \\ &= \frac{1}{2}\left(2^{6-k}-2^k \pm 2^{-k}|2^k-8|\sqrt{2^{2k} - 2^{k+2} + 64}\right) \\ &= \frac{64 - 2^{2k} \pm |2^k-8|\sqrt{2^{2k}-2^{k+2}+64}}{2^{k+1}} \\ a_3 = 5+x &= 5 + \frac{64 - 2^{2k} \pm |2^k-8|\sqrt{2^{2k}-2^{k+2}+64}}{2^{k+1}}\end{aligned}$

It's apparent that the maximum value will occur when the sign is '+'. Then we can just test all 7 cases for $k$ to find the maximum occurs for $k=0$ at $a_3 = \frac{73+7\sqrt{61}}{2}$

Therefore the answer is $a+b+c+d = 73 + 7 + 61 + 2 = \boxed{143.}$