6. A

Hint: $V_{1}=\frac{1}{6} \cdot B D \cdot C D \cdot A D \sin \theta$
$$
\begin{array}{l}
=\frac{1}{6} \cdot c \cos B \cdot b \cos C \cdot \frac{2 \triangle}{a} \cdot \sin \theta=\frac{\triangle}{3} \cdot \frac{b x \cos B \cos C \sin \theta}{a} \\
=\frac{\triangle \cos A \cos B \cos C \sin \theta}{3} \cdot \frac{b c}{a \cos A} \\
=\frac{\triangle \cos A \cos B \cos C \sin \theta}{3} \cdot \frac{2 b^{2} c^{2}}{a\left(-a^{2}+b^{2}+c^{2}\right)} \\
=\frac{2 a^{2} b^{2} c^{2} \cdot \triangle \cdot \cos A \cos B \cos C \sin \theta}{3} \cdot \frac{1}{a^{3}\left(-a^{2}+b^{2}+c^{2}\right)}
\end{array}
$$
(Above $\triangle$ represents the area of $\triangle A B C$).
Let $K=\frac{2 a^{2} b^{2} c^{2} \cdot \triangle \cdot \cos A \cos B \cos C \sin \theta}{3}$, similarly, we can get $V_{2}=\frac{K}{b^{3}\left(a^{2}-b^{2}+c^{2}\right)}, V_{3}=\frac{K}{c^{3}\left(a^{2}+b^{2}-c^{2}\right)}$
Since $K$ is the same value, to compare the sizes of $V_{1}, V_{2}, V_{3}$, it is sufficient to compare $a^{3}\left(-a^{2}+b^{2}+c^{2}\right), b^{3}\left(a^{2}-b^{2}+c^{2}\right)$, and $c^{3}\left(a^{2}+b^{2}-c^{2}\right)$.
Since $a>b, a^{2}+b^{2}>c^{2}$,
we have $a^{3}\left(-a^{2}+b^{2}+c^{2}\right)-b^{3}\left(a^{2}-b^{2}+c^{2}\right)=-\left(a^{3}-b^{3}\right)\left(a^{2}+b^{2}\right)+c^{2}\left(a^{3}-b^{3}\right)-2 a^{2} b^{2}(a-b)$
$$
=\left(a^{3}-b^{3}\right)\left(-a^{2}-b^{2}+c^{2}\right)-2 a^{2} b^{2}(a-b)\frac{1}{b^{3}\left(a^{2}-b^{2}+c^{2}\right)}$,
so $V_{1}>V_{2}$. Similarly, $V_{2}>V_{3}$. Therefore, $V_{1}>V_{2}>V_{3}$, choose A.
6. In an acute triangle $\triangle A B C$, $B C=a, C A=b, A B=c$, and $a>b>c$. By folding the triangle along the altitudes $A D, B E, C F$ of sides $B C, C A, A B$ respectively, the triangle is folded into dihedral angles all equal to $\theta(0V_{2}>V_{3}$
B. $V_{1}V_{3}>V_{2}$ or $V_{3}>V_{1}>V_{2}$
D. The sizes of $V_{1}, V_{2}, V_{3}$ cannot be determined