Đáp án đúng: D

Phương pháp giải:

- Kẻ \(OM \bot AC\,\,\left( {M \in AC} \right)\), \(ON \bot AB\,\,\left( {N \in AB} \right)\), \(OP \bot BC\,\,\left( {P \in BC} \right)\). Khi đó ta có \(OP = a,\) \(OM = a\sqrt 2 ,\) \(ON = a\sqrt 3 \).
- Trong \(\left( {OCN} \right)\) kẻ \(OH \bot CN\,\,\left( {H \in CN} \right)\), chứng minh \(OH \bot \left( {ABC} \right)\).
- Sử dụng hệ thức lượng trong tam giác vuông để tính khoảng cách.Giải chi tiết:
Kẻ \(OM \bot AC\,\,\left( {M \in AC} \right)\), \(ON \bot AB\,\,\left( {N \in AB} \right)\), \(OP \bot BC\,\,\left( {P \in BC} \right)\).
Khi đó ta có \(OP = a,\,\,OM = a\sqrt 2 ,\,\,ON = a\sqrt 3 \).
Trong \(\left( {OCN} \right)\) kẻ \(OH \bot CN\,\,\left( {H \in CN} \right)\) ta có:
\(\begin{array}{l}\left\{ \begin{array}{l}AB \bot ON\\AB \bot OC\end{array} \right. \Rightarrow AB \bot \left( {OCN} \right) \Rightarrow AB \bot OH\\\left\{ \begin{array}{l}OH \bot AB\\OH \bot CN\end{array} \right. \Rightarrow OH \bot \left( {ABC} \right) \Rightarrow d\left( {O;\left( {ABC} \right)} \right) = OH\end{array}\)
Áp dụng hệ thức lượng trong tam giác vuông ta có: \(\dfrac{1}{{O{H^2}}} = \dfrac{1}{{O{C^2}}} + \dfrac{1}{{O{N^2}}} = \dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{B^2}}} + \dfrac{1}{{O{C^2}}}\)
Lại có
\(\begin{array}{l}\dfrac{1}{{O{M^2}}} = \dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{C^2}}};\,\,\dfrac{1}{{O{N^2}}} = \dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{B^2}}};\,\,\dfrac{1}{{O{P^2}}} = \dfrac{1}{{O{B^2}}} + \dfrac{1}{{O{C^2}}}\\ \Rightarrow \dfrac{1}{{O{M^2}}} + \dfrac{1}{{O{N^2}}} + \dfrac{1}{{O{P^2}}} = 2\left( {\dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{B^2}}} + \dfrac{1}{{O{C^2}}}} \right)\\ \Rightarrow \dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{B^2}}} + \dfrac{1}{{O{C^2}}} = \dfrac{1}{2}\left( {\dfrac{1}{{O{M^2}}} + \dfrac{1}{{O{N^2}}} + \dfrac{1}{{O{P^2}}}} \right)\\ \Rightarrow \dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{B^2}}} + \dfrac{1}{{O{C^2}}} = \dfrac{1}{2}\left( {\dfrac{1}{{2{a^2}}} + \dfrac{1}{{3{a^2}}} + \dfrac{1}{{{a^2}}}} \right) = \dfrac{{11}}{{12{a^2}}}\\ \Rightarrow \dfrac{1}{{O{H^2}}} = \dfrac{{11}}{{12{a^2}}} \Rightarrow OH = \dfrac{{2a\sqrt {33} }}{{11}}\end{array}\)
Vậy \(d\left( {O;\left( {ABC} \right)} \right) = \dfrac{{2a\sqrt {33} }}{{11}}\).
Chọn D