6. The number of planes for which the ratio of distances to the 4 vertices of a regular tetrahedron is $1: 1: 1: 2$ is $\qquad$.
6.32

Let the vertices of a regular tetrahedron be $A, B, C, D$. The plane $\alpha$ that divides the distances to these four points in the ratio $1: 1: 1: 2$ can be of two types:
(1) $A, B, C$ are on the same side of $\alpha$, there are 2 such planes (as shown in Figure 8).
(1) $\frac{A A_{1}}{A_{1} D}=\frac{B B_{1}}{B_{1} D}=\frac{C C_{1}}{C_{1} D}=\frac{1}{2}$
(2) $\frac{A_{1} A}{A D}=\frac{B_{1} B}{B D}=\frac{C_{1} C}{C D}=1$
Figure 8
(2) $A, B, C$ are on opposite sides of $\alpha$, there are 6 such planes (as shown in Figure 9).
(1) $\frac{A A_{1}}{A_{1} D}=\frac{B B_{1}}{B_{1} D}=\frac{1}{2}$
(2) $\frac{C_{1}}{C_{1} D}=\frac{1}{2}$
(3) $\frac{A C_{1}}{C_{1} D}=\frac{1}{2}$
$\frac{A D_{1}}{D_{1} C}=\frac{B C_{1}}{C_{1} C}=1$
$\frac{A A_{1}}{A_{1} C}=\frac{B B_{1}}{B_{1} C}=1$
$\frac{A A_{1}}{A_{1} B}=\frac{A B_{1}}{B_{1} C}=1$
(4) $\frac{A A_{1}}{A_{1} B}=\frac{A B_{1}}{B_{1} C}=1$
(5) $\frac{A A_{1}}{A_{1} B}=\frac{B B_{1}}{B_{1} C}=1$
(6) $\frac{A A_{1}}{A_{1} B}=\frac{B B_{1}}{B_{1} C}=1$
$\frac{B D_{1}}{D_{1} D}=\frac{O C_{1}}{C_{1} D}=\frac{1}{2}$
$\frac{B C_{1}}{C_{1} D}=\frac{1}{2}$
$\frac{A D_{1}}{D_{1} D}=\frac{C C_{1}}{C_{1} D}=\frac{1}{2}$

By permuting $A, B, C, D$, a total of $8 \times 4=32$ planes can be obtained.