$3^a+3^b+3^c = 7299$

Find the total number of positive integers ordered pairs of the equation $3^a+3^b+3^c = 7299$.

WLOG, let $c>b>a$.

Rewrite the RHS of the given equation as the product of two factors, we have:

$7299 = 3^a + 3^b + 3^c$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,=3^a\left(1 + \dfrac{3^b}{3^a} +\dfrac{3^c}{3^a}\right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,=3^a(1 + 3^{b-a} + 3^{c-a})$

Note that $7299 = 3^2 \cdot 811$, thus $a$ is either 1 or 2.

But notice also that $(1 + 3^{b-a} + 3^{c-a})$ is not divisible by 3, hence, $3^2=3^a$ is the only solution for this case. This tells us $a=2$.

By substituting this back to the original equation gives

$3^2+3^b+3^c = 7299$

$3^b+3^c = 7299-9$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= 7290$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= 3^6(10)$

Again if we factor $3^b+3^c$ as the product of two factors, we see that

$3^b\left(1+ \dfrac{3^c}{3^b}\right) = 3^6(10)$

$3^b(1+ 3^{c-b}) = 3^6(10)$

As $1+ 3^{c-b}$ is not divisible by 3, we need $3^b=3^6\,\,\implies\,\,b=6$.

$\therefore 1+ 3^{c-6}=10$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3^{c-6}=9$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=3^2\,\,\,\implies\,\,c=8$

$(a,\,b,\,c) = (2,\,6,\,8)$ is the only possible solution up to rearrangement.

Hence the total number of positive integers ordered pairs that satisfy the given equation is $3!=6$.

In fact, $(a,\,b,\,c) = (2,\,6,\,8),\,(2,\,8,\,6),\,(6,\,2,\,8),\,(6,\,8,\,2),\,(8,\,2,\,6),\,(8,\,6,\,2)$.