Trong không gian $Oxyz$, xét số thực $m\in \left( 0;1 \right)$ và...

Gọi $I\left( a;b;c \right)$ là tâm mặt cầu.
Theo giả thiết ta có $R=d\left( I,\left( \alpha \right) \right)=d\left( I,\left( \beta \right) \right)$.
Mà $d\left( I,\left( \beta \right) \right)=\dfrac{\left| \dfrac{a}{m}+\dfrac{b}{1-m}+c-1 \right|}{\sqrt{\dfrac{1}{{{m}^{2}}}+\dfrac{1}{{{\left( 1-m \right)}^{2}}}+1}}$
Ta có
$\begin{aligned}
& \sqrt{\dfrac{1}{{{m}^{2}}}+\dfrac{1}{{{\left( 1-m \right)}^{2}}}+1}=\sqrt{{{\left( \dfrac{1}{m}+\dfrac{1}{1-m} \right)}^{2}}-2\dfrac{1}{m}.\dfrac{1}{1-m}+1} \\
& =\sqrt{{{\left[ \dfrac{1}{m\left( 1-m \right)} \right]}^{2}}-2\dfrac{1}{m}.\dfrac{1}{1-m}+1}=\dfrac{1}{m\left( 1-m \right)}-1(\text{do} m\in \left( 0;1 \right) \\
\end{aligned}$
Nên
$\begin{aligned}
& R=\dfrac{\left| \dfrac{a\left( 1-m \right)+bm+cm\left( 1-m \right)-m\left( 1-m \right)}{m\left( 1-m \right)} \right|}{\dfrac{1}{m\left( 1-m \right)}-1} \\
& \Leftrightarrow R=\dfrac{\left| a-am+bm+cm-c{{m}^{2}}-m+{{m}^{2}} \right|}{{{m}^{2}}-m+1} \\
& \Leftrightarrow \left[ \begin{aligned}
& R-Rm+R{{m}^{2}}=a-am+bm+cm-c{{m}^{2}}-m+{{m}^{2}} \\
& -R+Rm-R{{m}^{2}}=a-am+bm+cm-c{{m}^{2}}-m+{{m}^{2}} \\
\end{aligned} \right. \\
& \Leftrightarrow \left[ \begin{aligned}
& {{m}^{2}}\left( R+c-1 \right)+m\left( a-b-c-R+1 \right)+R-a=0\left( 1 \right) \\
& {{m}^{2}}\left( R+c-1 \right)+m\left( b+c-a-R-1 \right)+R+a=0\left( 2 \right) \\
\end{aligned} \right. \\
\end{aligned}$
Xét (1) do mặt cầu tiếp xúc với tiếp xúc đồng thời với cả hai mặt phẳng $\left( \alpha \right),\left( \beta \right)$ với mọi $m\in \left( 0;1 \right)$ nên pt (1) nghiệm đúng với mọi $m\in \left( 0;1 \right)$.
$\Leftrightarrow \left\{ \begin{aligned}
& R+c-1=0 \\
& a-b-c-R+1=0 \\
& R-a=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=R \\
& b=R \\
& c=1-R \\
\end{aligned} \right.\Rightarrow I\left( R;R;1-R \right)$.
Mà $R=d\left( I,\left( \alpha \right) \right)\Leftrightarrow R=\dfrac{\left| 2R-R+2\left( 1-R \right)+10 \right|}{3}\Leftrightarrow 3R=\left| 12-R \right|\Leftrightarrow \left[ \begin{aligned}
& R=3 \\
& R=-6(l) \\
\end{aligned} \right.$
Xét (2) tương tự ta được
$\Leftrightarrow \left\{ \begin{aligned}
& R+c-1=0 \\
& b+c-a-R-1=0 \\
& R+a=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-R \\
& b=-R \\
& c=R+1 \\
\end{aligned} \right.\Rightarrow I\left( -R;-R;R+1 \right)$
Mà $R=d\left( I,\left( \alpha \right) \right)\Leftrightarrow R=\dfrac{\left| -2R+R+2\left( 1+R \right)+10 \right|}{3}\Leftrightarrow 3R=\left| 12+R \right|\Leftrightarrow \left[ \begin{aligned}
& R=6 \\
& R=-3(l) \\
\end{aligned} \right.$.
Vậy ${{R}_{1}}+{{R}_{2}}=9$.