Phương pháp giải:

\(d\left( {B;\left( {SAD} \right)} \right) = \frac{{3{V_{S.ABD}}}}{{{S_{SAD}}}}\)

Lời giải chi tiết:

Ta có: \({S_{ABD}} = \frac{1}{2}{S_{ABCD}} \Rightarrow {V_{S.ABD}} = \frac{1}{2}{V_{S.ABCD}} = 3{a^3}\).

Ta có: \(\left\{ \begin{array}{l}AD \bot AB\\AD \bot SA\end{array} \right. \Rightarrow AD \bot \left( {SAB} \right) \Rightarrow AD \bot SA \Rightarrow \Delta SAD\) vuông tại A.

\(SB = \frac{{3{V_{S.ABCD}}}}{{{S_{ABCD}}}} = \frac{{18{a^3}}}{{3a.\sqrt {25{a^2} - 9{a^2}} }} = \frac{{18{a^3}}}{{3a.4a}} = \frac{{3a}}{2}\)

\( \Rightarrow SA = \sqrt {S{B^2} + A{B^2}}  = \sqrt {\frac{{9{a^2}}}{4} + 9{a^2}}  = \frac{{3a\sqrt 5 }}{2}\)

\( \Rightarrow {S_{SAD}} = \frac{1}{2}SA.AD = \frac{1}{2}.\frac{{3a\sqrt 5 }}{2}.4a = 3{a^2}\sqrt 5 \)

Vậy \(d\left( {B;\left( {SAD} \right)} \right) = \frac{{3{V_{S.ABD}}}}{{{S_{SAD}}}} = \frac{{3.3{a^3}}}{{3{a^2}\sqrt 5 }} = \frac{{3a\sqrt 5 }}{5}\).

Chọn A