Solution:
First we prove that the given equation has at least one root. To do this we consider the following function.

$$f(x) = 9^{x + 1} + 2187 - 3^{6x - x^{2}}$$

Note that this function is continuous. Note also that $f(0) = 9 + 2187 - 1 > 0$ and $f(3) = 9^{4} + 2187 - 3^{9} < 0$. Therefore by the intermediate value theorem there exists a real number $0 < z < 3$ such that $f(z) = 0$. Therefore there is at least one value $x$ which satisfies the equation.

Next, we can rewrite the equation in the following form

$$3^{x} + 3^{5 - x} = 3^{x(5 - x) - 2} \quad (1)$$

Let $r$ and $s$ be any two numbers such that $r + s = 5$. Notice that if $r$ is a root of (1) then $s$ must also be a root (and vice versa). This is because

$$3^{r} + 3^{5 - r} = 3^{5 - s} + 3^{s} \qquad \mathrm{and} \qquad 3^{r(5 - r) - 2} = 3^{(5 - s)s - 2}.$$

Now we claim that if $x_{0}$ is the smallest root of (1), then $(5 - x_{0})$ must be the largest root. Indeed if $x_{1}$ were a root of (1) greater than $(5 - x_{0})$, then $(5 - x_{1})$ would also have to be a root, but $(5 - x_{1}) < (5 - (5 - x_{0})) = x_{0}$ and this would contradict the fact that $x_{0}$ is the smallest root.

Therefore, the sum of the largest and the smallest roots of (1) is $x_{0} + (5 - x_{0}) = 5$