Ta có \({S_{ACD}} = \frac{1}{2}{S_{ABCD}} = \frac{3}{2}\).

\( \Rightarrow {V_{S.ACD}} = \frac{1}{3}SI \cdot {S_{ACD}} = \frac{1}{3} \cdot \frac{3}{2} \cdot \frac{3}{2} = \frac{3}{4}\).

Ta có: \(\left\{ {\begin{array}{*{20}{l}}{AD \bot AB}\\{AD \bot SI}\end{array}} \right. \Rightarrow AD \bot \left( {SAB} \right) \Rightarrow AD \bot SA\)

\( \Rightarrow \Delta SAD\)vuông tại \(A\).

Xét tam giác vuông \(SAI\) nên  \(SA = \sqrt {S{I^2} + A{I^2}}  = \sqrt {{{\left( {\frac{3}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}  = \sqrt 3 \)

\( \Rightarrow {S_{SAD}} = \frac{1}{2}SA \cdot AD = \frac{1}{2} \cdot \sqrt 3  \cdot \sqrt 3  = \frac{3}{2}\)

Vậy \(d\left( {C\,,\,\left( {SAD} \right)} \right) = \frac{{3{V_{S.ACD}}}}{{{S_{SAD}}}} = \frac}{{\frac{3}{2}}} = \frac{3}{2}.\)

Đáp án: \(\frac{3}{2}\).