Cho hình chóp đều S.ABCD có thể tích bằng 1/3 , đáy ABCD

Ta có $V_{S.ABC\text{D}}=\frac{1}{2}{S}_{ABC\text{D}}.d_{\left(S,\left(ABC\text{D}\right)\right)}\Rightarrow d_{\left(S,\left(ABC\text{D}\right)\right)}=1$

Đặt $\left(ABC\text{D}\right):ax+by+cz+d=0\left(a^2+b^2+c^2\ne 0\right)$ .

Vì $AB\subset \left(ABC\text{D}\right)$  nên ${\overrightarrow{n}}_{\left(ABC\text{D}\right)}\perp {\overrightarrow{u}}_{AB}\iff {\overrightarrow{n}}_{ABC\text{D}}.{\overrightarrow{u}}_{AB}=0\iff b=0$ .

Vì $M\left(1;0;1\right)\in \text{A}B\subset \left(ABC\text{D}\right)$  nên $a+c+d=0\Rightarrow d=-a-c$ .

Ta có: $d_{\left(S,\left(ABC\text{D}\right)\right)}=1\iff \frac{\left|-a-c\right|}{\sqrt{a^2+c^2}}=1\iff \left|a+c\right|=\sqrt{a^2+c^2}\iff 2\text{a}c=0\iff \left[\begin{array}{l}a=0\\ c=0\end{array}\right.$ .

Trường hợp 1: a=0. Chọn $c=1\Rightarrow \left(ABC\text{D}\right):z+1=0$ .

Trường hợp 2: c=0. Chọn $a=1\Rightarrow \left(ABC\text{D}\right):x+1=0$ .