6. Given that the hyperbola has asymptotes $2 x \pm y=0$, and passes through the intersection point of the lines $x+y-3=0$ and $2 x-y+3 t=0$, where $-2 \leqslant t \leqslant 5$. Then the maximum possible value of the real axis length of the hyperbola is $\qquad$
6. $4 \sqrt{3}$.

It is easy to know that the intersection point of the two lines is $(1-t, 2+t)$.
First, consider the case where the foci of the hyperbola are on the $x$-axis.
$$
\text { Let } \frac{x^{2}}{a^{2}}-\frac{y^{2}}{4 a^{2}}=1(a>0) \text {. }
$$

Notice that, $-2 \leqslant t \leqslant 5$. Then
$$
\begin{array}{l}
4 a^{2}=4(1-t)^{2}-(2+t)^{2} \\
=3(t-2)^{2}-12 \\
\leqslant 3 \times 4^{2}-12=36 .
\end{array}
$$

When $t=-2$, the maximum value of the real axis length $2 a$ is 6. Next, consider the case where the foci of the hyperbola are on the $y$-axis.
Let $\frac{y^{2}}{a^{2}}-\frac{4 x^{2}}{a^{2}}=1(a>0)$. Then
$$
\begin{array}{c}
a^{2}=(2+t)^{2}-4(1-t)^{2} \\
=-3(t-2)^{2}+12 \leqslant 12 .
\end{array}
$$

When $t=2$, the maximum value of the real axis length $2 a$ is $4 \sqrt{3}$.
Since $4 \sqrt{3}>6$, the maximum possible value of the real axis length of the hyperbola is $4 \sqrt{3}$.