Lời giải:.png)
Ta có \(BC=\sqrt{A{{C}^{2}}+A{{B}^{2}}-2AC.AB.cos{{120}^{0}}}=a\sqrt{7};\)
\(\begin{align}
& {{A}_{1}}B=\sqrt{{{A}_{1}}{{A}^{2}}+A{{B}^{2}}}=a\sqrt{21};{{A}_{1}}K=\sqrt{{{A}_{1}}{{C}_{1}}^{2}+{{C}_{1}}{{K}^{2}}}=3a,KB=\sqrt{K{{C}^{2}}+C{{B}^{2}}}=2a\sqrt{3} \\
& d(I,({{A}_{1}}BK))=\frac{1}{2}d\left( {{B}_{1}},\left( {{A}_{1}}BK \right) \right)=\frac{1}{2}.\frac{3{{V}_{{{B}_{1}}{{A}_{1BK}}}}}{{{S}_{\Delta {{A}_{1}}BK}}} \\
\end{align}\)
Mà \({{V}_{{{B}_{1}}{{A}_{1}}BK}}=\frac{1}{2}{{V}_{K.{{A}_{1}}{{B}_{1}}BA}}=\frac{1}{2}.\frac{2}{3}{{V}_{ABC.{{A}_{1}}{{B}_{1}}{{C}_{1}}}}=\frac{1}{3}.2a\sqrt{5}.\frac{1}{2}.a.2a.\sin {{120}^{0}}=\frac{{{a}^{3}}\sqrt{15}}{3}.\)
Theo công thức Herong, diện tích tam giác \({{A}_{1}}BK\) bằng
\(S=\sqrt{p\left( p-2a\sqrt{3} \right)\left( p-3a \right)\left( p-a\sqrt{21} \right)}=3{{a}^{2}}\sqrt{3}\) với \(p=\frac{2a\sqrt{3}+3a+a\sqrt{21}}{2}\)
Vậy \(d\left( I,\left( {{A}_{1}}BK \right) \right)=\frac{3}{2}.\frac{\frac{{{a}^{3}}\sqrt{15}}{3}}{3{{a}^{2}}\sqrt{3}}=\frac{a\sqrt{5}}{6}.\)
