Can you make it the best?

We begin like this:

$3^n(a + b) = 2^nab$

$a| 3^n b , b| 3^na \to a = 3^{n_1} x, b = 3^{n_2}x , 3 \nmid x$

$3^n x(3^{n_1} + 3^{n_2}) = 2^n x^2 3^{n_1 + n_2}$

Let $n_1 \geq n_2 \geq 0$

$3^n (3^{n_1-n_2} + 1) = 2^n x 3^{n_1} \to n = n_1$

$3^{n - n_2} + 1 = 2^n x$

$1 + 3^N mod 8 = 4, 2$

$n = 1$

$1 + 3^{1-n_2} = 2x$

$n_2 = 0 \to x = 2,a = 6,b = 2$

$n_2 = 1 \to x = 1,a = 3,b = 3$

$n = 2$

$1 + 3^{2-n_2} = 4x$

$n_2 = 1 \to x = 1, a = 9,b = 3$

Triples are $(a,b,n) = (6,2,1), (3,3,1), (9,3,2)$ and its permutations.

Maximum value = $9 + 3 + 2 = 14$ .