Trong không gian với hệ trục $Oxyz$, một mặt phẳng đi qua điểm...

Gọi $A\left( a;0;0 \right),B\left( 0;b;0 \right),C\left( 0;0;c \right)$ với $a,b,c>0$
Phương trình mặt phẳng $\left( ABC \right):\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$
$M\left( 16;4;9 \right)\in \left( ABC \right)\Leftrightarrow \dfrac{16}{a}+\dfrac{4}{b}+\dfrac{9}{c}=1 (1)$
$P=4OA+OB+OC=4a+b+c$
$P=\left( 4a+b+c \right).1=\left( 4a+b+c \right).\left( \dfrac{16}{a}+\dfrac{4}{b}+\dfrac{9}{c} \right)=77+16\left( \dfrac{a}{b}+\dfrac{b}{a} \right)+4\left( \dfrac{4c}{a}+\dfrac{9a}{c} \right)+\left( \dfrac{4c}{b}+\dfrac{9b}{c} \right)$
$P\overset{Cauchy}{\mathop{\ge }} 77+16.2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+4.2\sqrt{\dfrac{4c}{a}.\dfrac{9a}{c}}+2\sqrt{\dfrac{4c}{b}.\dfrac{9b}{c}}=169$
$\min P\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{a}{b}=\dfrac{b}{a} \\
& \dfrac{4c}{a}=\dfrac{9a}{c} \\
& \dfrac{4c}{b}=\dfrac{9b}{c} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=b \\
& c=\dfrac{3}{2}a \\
\end{aligned} \right.$
Thế vào (1), ta có: $\dfrac{16}{a}+\dfrac{4}{a}+\dfrac{9}{\dfrac{3}{2}a}=1\Leftrightarrow b=a=26;c=39$
Thể tích khối chóp $OABC$ là: $V=\dfrac{1}{6}OA.OB.OC=4394$