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设棱锥S-ABC高SH,H是正三角形ABC的外心(内心、重心),连结AH与BC相交于M,
连结SM,AM=√3BC/2=(√3/2)*2√6=3√2,HM=AM/3=√2,
SM=√(MH^2+SH^2)=√3,S△SBC=SM*BC/2=√3*2√6/2=3√2,
S△ABC=√3/4(2√6)^2=6√3,
VS-ABC=S△ABC*SH/3=2√3,
VS-ABC=(S△ABC+S△SBC+S△SAC+S△SAB)*R/3=(6√3+3*3√2)R/3,
(6√3+3*3√2)R/3=2√3,
∴R=√6-2.