If $\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{9^x-2 \cdot 3^x+1}{\log (1+3 x) \cdot \tan 2 x} & , \text { if } x \neq 0 \\ \mathrm{a}(\log \mathrm{b})^{\mathrm{c}} & , \text { if } x=0\end{array}\right.$ is continuous at $x=0$, then $\mathrm{a}+\mathrm{b}+\mathrm{c}=$

To ensure continuity at $x = 0$, the limit must satisfy $\lim_{x \to 0} f(x) = f(0)$.

Direct evaluation gives $f(0) = a (\log b)^c$.

The limit expression is $\lim_{x \to 0} \frac{9^x - 2 \cdot 3^x + 1}{\log(1 + 3x) \cdot \tan 2x}$.
Factor the numerator: $9^x - 2 \cdot 3^x + 1 = (3^x - 1)^2$.

Rewriting the limit: $\lim_{x \to 0} \frac{(3^x - 1)^2}{\log(1 + 3x) \cdot \tan 2x}$.

Using known limits $\lim_{x \to 0} \frac{a^x - 1}{x} = \log a$, $\lim_{x \to 0} \frac{\log(1 + kx)}{x} = k$, and $\lim_{x \to 0} \frac{\tan kx}{x} = k$, divide by $x^2$:

$\lim_{x \to 0} \frac{\left(\frac{3^x - 1}{x}\right)^2}{\left(\frac{\log(1 + 3x)}{x}\right) \cdot \left(\frac{\tan 2x}{x}\right)} = \frac{(\log 3)^2}{3 \cdot 2} = \frac{(\log 3)^2}{6}$.

Setting $f(0) = \lim_{x \to 0} f(x)$ yields $a (\log b)^c = \frac{(\log 3)^2}{6}$.

Matching terms gives $a = \frac{1}{6}$, $b = 3$, and $c = 2$.

The sum is $a + b + c = \frac{1}{6} + 3 + 2 = \frac{31}{6}$.

$\boxed{\frac{31}{6}}$