Chọn B
Thể tích khối tứ diện ABCD là
\(\[V_{ABCD} =\frac{1}{6} AB.AC.AD\sqrt{1-\cos ^{2} \widehat{BAC}-\cos ^{2} \widehat{BAD}-\cos ^{2} \widehat{CAD}+2\cos \widehat{BAC}\cos \widehat{BAD}\cos \widehat{CAD}} \] \)
\(\[=\frac{1}{6} 3.4.6\sqrt{1-\frac{1}{4} -\frac{1}{4} -0+0} =6\sqrt{2} .\] \)
Mặt khác lại có
\(\[V_{ABCD} =\frac{1}{6} AB.CD.d\left(AB,CD\right)\sin \left(AB,CD\right).\] \)
Ta có
\(● CD=\sqrt{AC^{2} +AD^{2} } =\sqrt{16+36} =2\sqrt{13} .\)
\(● \cos \left(\overrightarrow{AB},\overrightarrow{CD}\right)=\frac{\overrightarrow{AB}.\overrightarrow{CD}}{AB.CD} =\frac{\overrightarrow{AB}.\left(\overrightarrow{AD}-\overrightarrow{AC}\right)}{AB.CD}\)
\(=\frac{\overrightarrow{AB}.\overrightarrow{AD}-\overrightarrow{AB}.\overrightarrow{AC}}{AB.CD}
\[=\frac{AB.AD.\cos 60{}^\circ -AB.AC\cos 60{}^\circ }{AB.CD} =\frac{3.6.\frac{1}{2} -3.4.\frac{1}{2} }{3.2\sqrt{13} } =\frac{1}{2\sqrt{13} } .\] \)
\(\[\Rightarrow \cos \left(AB,CD\right)=\frac{1}{2\sqrt{13} } \Rightarrow \sin \left(AB,CD\right)=\sqrt{1-\left(\frac{1}{2\sqrt{13} } \right)^{2} } =\frac{\sqrt{51} }{2\sqrt{13} } .\] \)
Khoảng cách giữa AB và CD là
\(\[d\left(AB,CD\right)=\frac{6V}{AB.CD.\sin \left(AB,CD\right)} =\frac{36\sqrt{2} }{3.2\sqrt{13} .\frac{\sqrt{51} }{2\sqrt{13} } } =\frac{4\sqrt{102} }{17} .\] \)