Let $t=3^x$. The intersection equation should become a quadratic equation in $t$.

If the intersection points have $x$-coordinates differing by $2$, determine the ratio of their corresponding $t$-values. Then use the sum and product of the quadratic's roots.

Algebra is faster here because the condition about the difference between the two $x$-coordinates becomes a simple ratio after setting $t=3^x$.

Graph $y=x^2-3x+0.81$. Click the two $x$-intercepts; they are $0.3$ and $2.7$.

Verify that $2.7\div0.3=9=3^2$. Therefore, the corresponding powers of $3$ have exponents that differ by $2$, matching the condition in the question.

At an intersection, $3^{x+1}+k=9^x+2k$. Let $t=3^x$. Then $3^{x+1}=3t$ and $9^x=(3^x)^2=t^2$, so the equation becomes $t^2-3t+k=0$.

Let the two positive roots be $t_1$ and $t_2$, with $t_2>t_1$. Since the corresponding $x$-coordinates differ by $2$, $\frac{t_2}{t_1}=\frac{3^{x_2}}{3^{x_1}}=3^2=9$. Thus, the roots can be written as $u$ and $9u$.

For $t^2-3t+k=0$, the roots have sum $3$ and product $k$. Therefore, $u+9u=3$, so $u=\frac{3}{10}$. The product is $u(9u)=9\left(\frac{3}{10}\right)^2=\frac{81}{100}$.