Gọi $A(a;0;0), B(0;b;0), C(0;0;c)$ với $a,b,c \ne 0$.

PT mặt phẳng $(\alpha)$ theo đoạn chắn: $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.

$M(1;9;4) \in (\alpha) \Rightarrow \frac{1}{a}+\frac{9}{b}+\frac{4}{c}=1$ (1).

Theo công thức khoảng cách:

$OA = \sqrt{a^2+0+0} = |a|$

$OB = \sqrt{0+b^2+0} = |b|$

$OC = \sqrt{0+0+c^2} = |c|$

Theo đề: $OA=OB=OC \Rightarrow |a|=|b|=|c| \ne 0 \Rightarrow b = \pm a$ và $c = \pm a$.

TH1: $b=a, c=a$.

(1) $\Rightarrow \frac{1}{a}+\frac{9}{a}+\frac{4}{a}=1 \Rightarrow \frac{14}{a}=1 \Rightarrow a=14$.

PT $(\alpha)$: $\frac{x}{14}+\frac{y}{14}+\frac{z}{14}=1 \Leftrightarrow x+y+z-14=0$.

TH2: $b=a, c=-a$.

(1) $\Rightarrow \frac{1}{a}+\frac{9}{a}-\frac{4}{a}=1 \Rightarrow \frac{6}{a}=1 \Rightarrow a=6$.

PT $(\alpha)$: $\frac{x}{6}+\frac{y}{6}+\frac{z}{-6}=1 \Leftrightarrow x+y-z-6=0$.

TH3: $b=-a, c=a$.

(1) $\Rightarrow \frac{1}{a}-\frac{9}{a}+\frac{4}{a}=1 \Rightarrow \frac{-4}{a}=1 \Rightarrow a=-4$.

PT $(\alpha)$: $\frac{x}{-4}+\frac{y}{4}+\frac{z}{-4}=1 \Leftrightarrow x-y+z+4=0$.

TH4: $b=-a, c=-a$.

(1) $\Rightarrow \frac{1}{a}-\frac{9}{a}-\frac{4}{a}=1 \Rightarrow \frac{-12}{a}=1 \Rightarrow a=-12$.

PT $(\alpha)$: $\frac{x}{-12}+\frac{y}{12}+\frac{z}{12}=1 \Leftrightarrow x-y-z+12=0$.

Vậy có 4 mặt phẳng thỏa mãn: $x+y+z-14=0$; $x+y-z-6=0$; $x-y+z+4=0$; $x-y-z+12=0$.