2. Given a positive geometric sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{6}+a_{5}+a_{4}-a_{3}-a_{2}-a_{1}=49 \text {. }
$$

Then the minimum value of $a_{9}+a_{8}+a_{7}$ is $\qquad$
2. 196.

Let the common ratio be $q$. From the condition, we have
$$
\left(q^{3}-1\right)\left(a_{3}+a_{2}+a_{1}\right)=49.
$$

Clearly, $q^{3}-1>0$.
Then $a_{3}+a_{2}+a_{1}=\frac{49}{q^{3}-1}$.
Thus, $a_{9}+a_{8}+a_{7}=q^{6}\left(a_{3}+a_{2}+a_{1}\right)$
$$
\begin{array}{l}
=\frac{49 q^{6}}{q^{3}-1}=49\left(\sqrt{q^{3}-1}+\frac{1}{\sqrt{q^{3}-1}}\right)^{2} \\
\geqslant 49 \times 4=196,
\end{array}
$$

When $\sqrt{q^{3}-1}=1$, i.e., $q=\sqrt[3]{2}$, the equality holds. Therefore, the minimum value of $a_{9}+a_{8}+a_{7}$ is 196.