Chọn phương án B.

Ta có \(\overrightarrow{AB}=(1;-1;2)\), \(\overrightarrow{AC}=(2;0;1)\).

Mặt phẳng \((ABC)\) đi qua điểm \(A(1;2;1)\) và nhận \(\left[\overrightarrow{AB},\overrightarrow{AC}\right]=(-1;3;2)\) làm vectơ pháp tuyến.

\(\begin{aligned}\Rightarrow(ABC)\colon&-(x-1)+3(y-2)+2(z-1)=0\\ \Leftrightarrow&-x+3y+2z-7=0.\end{aligned}\)

Khi đó: $$\begin{aligned}DH&=d\left(D,(ABC)\right)\\ &=\dfrac{\left|-1+3\cdot1+2\cdot1-7\right|}{\sqrt{(-1)^2+3^2+2^2}}\\ &=\dfrac{3\sqrt{14}}{14}.\end{aligned}$$

Chọn phương án B.

Ta có \(\overrightarrow{AB}=(1;-1;2)\), \(\overrightarrow{AC}=(2;0;1)\), \(\overrightarrow{AD}=(0;-1;0)\). Khi đó:

• \(\left[\overrightarrow{AB},\overrightarrow{AC}\right]=(-1;3;2)\)
  \(\begin{aligned}\Rightarrow S_{ABC}&=\dfrac{1}{2}\left|\left[\overrightarrow{AB},\overrightarrow{AC}\right]\right|\\ &=\dfrac{\sqrt{(-1)^2+3^2+2^2}}{2}\\ &=\dfrac{\sqrt{14}}{2}\end{aligned}\)
• \(\left[\overrightarrow{AB},\overrightarrow{AC}\right]\cdot\overrightarrow{AD}=-1\cdot0+3\cdot(-1)+2\cdot0=-3\)
  \(\begin{aligned}\Rightarrow V_{D.ABC}&=\dfrac{1}{6}\left|\left[\overrightarrow{AB},\overrightarrow{AC}\right]\cdot\overrightarrow{AD}\right|\\ &=\dfrac{1}{2}\end{aligned}\)

Theo đề bài ta có: $$\begin{eqnarray*}
V_{ABCD}&=&\dfrac{1}{3}\cdot S_{ABC}\cdot DH\\
\Rightarrow DH&=&\dfrac{3V_{ABCD}}{S_{ABC}}\\
&=&\dfrac{3\cdot\dfrac{1}{2}}{\dfrac{\sqrt{14}}{2}}=\dfrac{3\sqrt{14}}{14}.
\end{eqnarray*}$$