$$
\begin{array}{l}
\text { 6. Let }(1+2 \sqrt{2}+3 \sqrt{3})^{n} \\
=a_{n}+b_{n} \sqrt{2}+c_{n} \sqrt{3}+d_{n} \sqrt{6} \text {, }
\end{array}
$$

where, $a_{n}, b_{n}, c_{n}, d_{n} \in \mathbf{N}$.
Then $\lim _{n \rightarrow+\infty} \frac{d_{n}}{a_{n}}=$ $\qquad$
$$
\begin{array}{l}
\text { 6. } \frac{\sqrt{6}}{6} . \\
\text { Let } \lambda_{1}=1+2 \sqrt{2}+3 \sqrt{3}, \\
\lambda_{2}=1-2 \sqrt{2}+3 \sqrt{3}, \\
\lambda_{3}=1+2 \sqrt{2}-3 \sqrt{3}, \\
\lambda_{4}=1-2 \sqrt{2}-3 \sqrt{3} .
\end{array}
$$

Then $\left.\left|\lambda_{1}\right|>\max || \lambda_{2}|,| \lambda_{3}|,| \lambda_{4} \mid\right\}$.
Hence $\lim _{n \rightarrow+\infty} \frac{\lambda_{i}^{n}}{\lambda_{1}^{n}}=0(i=2,3,4)$.
$$
\begin{array}{l}
\text { Also } \lambda_{1}^{n}=a_{n}+b_{n} \sqrt{2}+c_{n} \sqrt{3}+d_{n} \sqrt{6}, \\
\lambda_{2}^{n}=a_{n}-b_{n} \sqrt{2}+c_{n} \sqrt{3}-d_{n} \sqrt{6}, \\
\lambda_{3}^{n}=a_{n}+b_{n} \sqrt{2}-c_{n} \sqrt{3}-d_{n} \sqrt{6}, \\
\lambda_{4}^{n}=a_{n}-b_{n} \sqrt{2}-c_{n} \sqrt{3}+d_{n} \sqrt{6} . \\
\text { Then } \frac{1}{4}=\lim _{n \rightarrow+\infty} \frac{\lambda_{1}^{n}+\lambda_{2}^{n}+\lambda_{3}^{n}+\lambda_{4}^{n}}{4 \lambda_{1}^{n}}=\lim _{n \rightarrow+\infty} \frac{a_{n}}{\lambda_{1}^{n}},
\end{array}
$$

and $\frac{1}{4}=\lim _{n \rightarrow+\infty} \frac{\lambda_{1}^{n}+\lambda_{4}^{n}-\lambda_{2}^{n}-\lambda_{3}^{n}}{4 \lambda_{1}^{n}}=\lim _{n \rightarrow+\infty} \frac{\sqrt{6} d_{n}}{\lambda_{1}^{n}}$.
Therefore, $\lim _{n \rightarrow+\infty} \frac{d_{n}}{a_{n}}=\frac{\sqrt{6}}{6}$.