Trong không gian cho Oxyz, $A\left(a;0;0\right), B\left(0;b;0\right), C\left(0;0;c\right)$, $D\left(a+a\sqrt{b^2+c^2};b\sqrt{a^2+c^2};c\sqrt{a^2+b^2}\right)$ (a>0, b>0, c>0). Diện tích tam giác ABC bằng $\frac{\sqrt{3}}{2}.$ Tìm khoảng cách từ B đến mặt phẳng (ACD) khi $V_{A.BCD}$ đạt giá trị lớn nhất.

Chọn A

$$\begin{gathered} \overrightarrow{AB}=\left(-a;b;0\right), \overrightarrow{AC}=\left(a,0,c\right), \overrightarrow{AD}=\left(a\sqrt{b^2+c^2};b\sqrt{a^2+c^2};c\sqrt{a^2+b^2}\right) \\ \left[\overrightarrow{AB},\overrightarrow{AC}\right]=\left(\left|\begin{array}{cc}b& 0\\ 0& c\end{array}\right|;\left|\begin{array}{cc}0& -a\\ c& -a\end{array}\right|;\left|\begin{array}{cc}-a& b\\ -a& 0\end{array}\right|\right)=\left(bc;ac;ab\right) \end{gathered}$$.
Vì diện tích tam giác bằng $\frac{\sqrt{3}}{2}$ nên:
$$\begin{gathered} S_{\Delta ABC}=\frac{\sqrt{3}}{2}\iff \frac{1}{2}\left|\left[\overrightarrow{AB},\overrightarrow{AC}\right]\right|=\frac{\sqrt{3}}{2}\iff \frac{1}{2}\sqrt{{\left(ab\right)}^2+{\left(bc\right)}^2+{\left(ac\right)}^2}=\frac{\sqrt{3}}{2} \\ \iff {\left(ab\right)}^2+{\left(bc\right)}^2+{\left(ac\right)}^2=3 \end{gathered}$$.
Thể tích của tứ diện ABCD là:
$$\begin{gathered} V_{ABCD}=\frac{1}{6}\left|\left[\overrightarrow{AB},\overrightarrow{AC}\right].\overrightarrow{AD}\right|=\frac{1}{6}\left|abc\sqrt{b^2+c^2}+abc\sqrt{a^2+c^2}+abc\sqrt{a^2+b^2}\right| \\ =\frac{1}{6}\left|bc\sqrt{a^2{b}^2+a^2{c}^2}+ac\sqrt{a^2{b}^2+b^2{c}^2}+ab\sqrt{a^2{c}^2+b^2{c}^2}\right| \end{gathered}$$

Áp dụng bất đẳng thức Bunhiacopxki: ${(bc\sqrt{a^2{b}^2+a^2{c}^2}+ac\sqrt{a^2{b}^2+b^2{c}^2}+ab\sqrt{a^2{c}^2+b^2{c}^2})}^2$
 $$\begin{gathered} \le \text{[}{\left(bc\right)}^2+{\left(ac\right)}^2+{\left(ab\right)}^2\text{](}{a}^2{b}^2+a^2{c}^2+a^2{b}^2+b^2{c}^2+a^2{c}^2+b^2{c}^2) \\ \iff {(bc\sqrt{a^2{b}^2+a^2{c}^2}+ac\sqrt{a^2{b}^2+b^2{c}^2}+ab\sqrt{a^2{c}^2+b^2{c}^2})}^2\le 2{\text{[}{\left(bc\right)}^2+{\left(ac\right)}^2+{\left(ab\right)}^2\text{]}}^2 \\ \iff {(bc\sqrt{a^2{b}^2+a^2{c}^2}+ac\sqrt{a^2{b}^2+b^2{c}^2}+ab\sqrt{a^2{c}^2+b^2{c}^2})}^2\le {2.3}^2 \\ \iff {(bc\sqrt{a^2{b}^2+a^2{c}^2}+ac\sqrt{a^2{b}^2+b^2{c}^2}+ab\sqrt{a^2{c}^2+b^2{c}^2})}^2\le 18 \\ \iff \left|bc\sqrt{a^2{b}^2+a^2{c}^2}+ac\sqrt{a^2{b}^2+b^2{c}^2}+ab\sqrt{a^2{c}^2+b^2{c}^2}\right|\le 3\sqrt{2} \end{gathered}$$

$V_{A.BCD}\le \frac{3\sqrt{2}}{6}$ hay $V_{A.BCD}\le \frac{\sqrt{2}}{2}$.
nên $\max V_{A.BCD}=\frac{\sqrt{2}}{2}$. Dấu "=" xảy ra khi và chỉ khi $a=b=c=1$.
Ta có: $\overrightarrow{AC}=\left(-1;0;1\right),\overrightarrow{AD}=\left(\sqrt{2};\sqrt{2};\sqrt{2}\right)$.
Nên: $\left[\overrightarrow{AC},\overrightarrow{AD}\right]=\left(\left|\begin{array}{cc}0& 1\\ \sqrt{2}& \sqrt{2}\end{array}\right|;\left|\begin{array}{cc}1& -1\\ \sqrt{2}& \sqrt{2}\end{array}\right|;\left|\begin{array}{cc}-1& 0\\ \sqrt{2}& \sqrt{2}\end{array}\right|\right)=\left(-\sqrt{2};2\sqrt{2};-\sqrt{2}\right)$.
Do đó: $S_{\Delta ACD}=\frac{1}{2}\left|\left[\overrightarrow{AC},\overrightarrow{AD}\right]\right|=\frac{1}{2}\sqrt{12}=\sqrt{3}$.
Vậy $d(B,(ACD\left)\right)=\frac{3V_{A.BCD}}{S_{\Delta ACD}}=\frac{3.\frac{\sqrt{2}}{2}}{\sqrt{3}}=\frac{\sqrt{6}}{2}$.