Chọn phương án A.

Vì các đường thẳng \(AB\), \(AD\), \(AS\) đôi một vuông góc nên ta đặt hình chóp \(S.ABCD\) vào hệ trục tọa độ \(Oxyz\) sao cho \(A\equiv O(0;0;0)\), \(B\in Ox\), \(D\in Oy\), \(S\in Oz\).

• \(\overrightarrow{AM}=\left(\dfrac{1}{2};0;\dfrac{1}{2}\right)\)
• \(\overrightarrow{AN}=\left(0;\dfrac{2}{3};\dfrac{1}{3}\right)\)
• \(\overrightarrow{AC}=(1;1;0)\)
• \(\left[\overrightarrow{AM},\overrightarrow{AN}\right]=\left(-\dfrac{1}{3};-\dfrac{1}{6};\dfrac{1}{3}\right)\)
• \(\left[\overrightarrow{AM},\overrightarrow{AN}\right]\cdot\overrightarrow{AC}=-\dfrac{1}{3}-\dfrac{1}{6}=-\dfrac{1}{2}\)

Suy ra \(\dfrac{1}{6}\left|\left[\overrightarrow{AM},\overrightarrow{AN}\right]\cdot\overrightarrow{AC}\right|=\dfrac{1}{6}\cdot\dfrac{1}{2}=\dfrac{1}{12}\).

Vậy \(V_{ACMN}=\dfrac{a^3}{12}\).

Chọn phương án A.

Ta có \(V_{S.ABCD}=\dfrac{1}{3}S_{ABCD}\cdot SA=\dfrac{1}{3}a^2\cdot a=\dfrac{a^3}{3}\).

Ngoài ra, vì \(ABCD\) là hình vuông cạnh \(a\) nên $$\begin{aligned}V_{S.ABC}=V_{S.ACD}=V_{S.ABD}=V_{S.BCD}&=\dfrac{1}{2}V_{S.ABCD}\\ &=\dfrac{a^3}{6}\end{aligned}$$

\(\blacksquare\) Xét hình chóp \(S.ABC\) ta có $$\begin{aligned}
\dfrac{V_{S.AMC}}{V_{S.ABC}}&=\dfrac{SA}{SA}\cdot\dfrac{SM}{SB}\cdot\dfrac{SC}{SC}=\dfrac{1}{2}\\
\Rightarrow V_{S.AMC}&=\dfrac{1}{2}V_{S.ABC}\\
\Rightarrow V_{MABC}&=\dfrac{1}{2}V_{S.ABC}=\dfrac{1}{2}\cdot\dfrac{a^3}{6}=\dfrac{a^3}{12}\;(1)
\end{aligned}$$
\(\blacksquare\) Xét hình chóp \(S.ABD\) ta có $$\begin{aligned}
\dfrac{V_{S.AMN}}{V_{S.ABD}}&=\dfrac{SA}{SA}\cdot\dfrac{SM}{SB}\cdot\dfrac{SN}{SD}=\dfrac{1}{2}\cdot\dfrac{2}{3}=\dfrac{1}{3}\\
\Rightarrow V_{SAMN}&=\dfrac{1}{3}V_{S.ABD}=\dfrac{1}{3}\cdot\dfrac{a^3}{6}=\dfrac{a^3}{18}\;(2)
\end{aligned}$$
\(\blacksquare\) Xét hình chóp \(S.ACD\) ta có $$\begin{aligned}
\dfrac{V_{S.ACN}}{V_{S.ACD}}&=\dfrac{SA}{SA}\cdot\dfrac{SC}{SC}\cdot\dfrac{SN}{SD}=\dfrac{2}{3}\\
\Rightarrow V_{S.ACN}&=\dfrac{2}{3}V_{S.ABC}\\
\Rightarrow V_{NACD}&=\dfrac{1}{3}V_{S.ACD}=\dfrac{1}{3}\cdot\dfrac{a^3}{6}=\dfrac{a^3}{18}\;(3)
\end{aligned}$$
\(\blacksquare\) Xét hình chóp \(S.BCD\) ta có $$\begin{aligned}
\dfrac{V_{S.CMN}}{V_{S.CBD}}&=\dfrac{SC}{SC}\cdot\dfrac{SM}{SB}\cdot\dfrac{SN}{SD}=\dfrac{1}{2}\cdot\dfrac{2}{3}=\dfrac{1}{3}\\
\Rightarrow V_{SCMN}&=\dfrac{1}{3}V_{S.CBD}=\dfrac{1}{3}\cdot\dfrac{a^3}{6}=\dfrac{a^3}{18}\;(4)
\end{aligned}$$