Trong không gian Oxyz, cho tứ diện ABCD có điểm A(1;1;1)B(2;0;2)C(-1;-1;0),

Áp dụng bất đẳng thức Cô-si ta có: $4=\frac{AB}{A{B}^{\text{'}}}+\frac{AC}{A{C}^{\text{'}}}+\frac{A\text{D}}{A}\ge 3\sqrt[3]{\frac{AB.AC.AD}{AB\text{'}.AC\text{'}.AD\text{'}}}$.

$\Rightarrow \frac{AB\text{'}.AC\text{'}.AD\text{'}}{AB.AC.AD}\ge \frac{27}{64}\Rightarrow \frac{V_{AB\text{'}C\text{'}D\text{'}}}{V_{ABCD}}=\frac{AB\text{'}.AC\text{'}.AD\text{'}}{AB.AC.AD}\ge \frac{27}{64}\Rightarrow V_{A{B}^{\text{'}}{C}^{\text{'}}{D}^{\text{'}}}\ge \frac{27}{64}{V}_{ABC\text{D}}$

Để $V_{A{B}^{\text{'}}{C}^{\text{'}}{D}^{\text{'}}}$ nhỏ nhất khi và chỉ khi $\frac{A{B}^{\text{'}}}{AB}=\frac{A{C}^{\text{'}}}{AC}=\frac{A}{A\text{D}}=\frac{3}{4}\Rightarrow \left\{\begin{array}{l}\overrightarrow{A{B}^{\text{'}}}=\frac{3}{4}\overrightarrow{AB}\Rightarrow B^{\text{'}}\left(\frac{7}{4};\frac{1}{4};\frac{7}{4}\right)\\ \left(B^{\text{'}}{C}^{\text{'}}{D}^{\text{'}}\right)\text{ // }\left(BC\text{D}\right)\end{array}\right.$.

Ta có $\overrightarrow{CB}=\left(3;1;2\right),\overrightarrow{C\text{D}}=\left(1;4;4\right)$ suy ra mặt phẳng (BCD) có véctơ pháp tuyến là

$\overrightarrow{n}=\left[\overrightarrow{CB};\overrightarrow{C\text{D}}\right]=\left(-4;-10;11\right)$.

Lúc đó mặt phẳng $\left(B^{\text{'}}{C}^{\text{'}}{D}^{\text{'}}\right)$ song song với mặt phẳng $\left(BC\text{D}\right)$ và đi qua $B^{\text{'}}\left(\frac{7}{4};\frac{1}{4};\frac{7}{4}\right)$

$\Rightarrow \left(B^{\text{'}}{C}^{\text{'}}{D}^{\text{'}}\right):16\text{x}+40y-44\text{z}+39=0$.