6. Given a regular tetrahedron $\mathrm{ABCD}$ and a point $\mathrm{P}$ inside it, such that $P A=P B=\sqrt{11}, P C=P D=\sqrt{17}$. Then the edge length of the tetrahedron $\mathrm{ABCD}$ is

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6.6

Analysis: Since $P A=P B, P C=P D$, point $P$ lies on the perpendicular bisector planes of segments $A B$ and $C D$, i.e., point $P$ lies on the common perpendicular segment of segments $A B$ and $C D$. Let the edge length be $2 a$, and the distance from point $P$ to segment $A B$ be $x$, then
$$
a^{2}+x^{2}=11,(\sqrt{2} a-x)^{2}+a^{2}=17
$$

Solving yields: $a=3$, hence the edge length is 6.