Trong không gian $Oxyz,$ cho $A\left( 1;2;4 \right),B\left( 0;0;1...

Mặt cầu có $I\left( -1;1;0 \right)$, $R=2.$ Gọi $r$ là bán kính đường tròn giao tuyến.
Xét $d=d\left( I;\left( P \right) \right)=\dfrac{\left| -a+b+3 \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\Rightarrow r=\sqrt{{{R}^{2}}-{{d}^{2}}}=\sqrt{4-{{d}^{2}}}\Rightarrow {{r}_{\min }}\Leftrightarrow {{d}^{2}}_{\max }.$
Từ $\left\{ \begin{array}{*{35}{l}}
A\in \left( P \right) \\
B\in \left( P \right) \\
\end{array} \right.\Rightarrow \left\{ \begin{array}{*{35}{l}}
a+2b+4c+3=0 \\
c+3=0 \\
\end{array} \right.\Rightarrow \left\{ \begin{array}{*{35}{l}}
c=-3 \\
a+2b=9 \\
\end{array} \right.\Rightarrow \left\{ \begin{array}{*{35}{l}}
c=-3 \\
a=9-2b \\
\end{array} \right.$
$\Rightarrow {{d}^{2}}=\dfrac{{{\left( 2b-9+b+3 \right)}^{2}}}{{{\left( 2b-9 \right)}^{2}}+{{b}^{2}}+9}=\dfrac{{{\left( 3b-6 \right)}^{2}}}{5{{b}^{2}}-36b+90}=9.\dfrac{{{\left( b-2 \right)}^{2}}}{5{{b}^{2}}-36b+90}=f\left( b \right)$
$\Rightarrow {f}'\left( b \right)=9.\dfrac{2\left( b-2 \right)\left( 5{{b}^{2}}-36b+90 \right)-{{\left( b-2 \right)}^{2}}\left( 10b-36 \right)}{{{\left( 5{{b}^{2}}-36b+90 \right)}^{2}}}=0$
$\Rightarrow 2\left( 5{{b}^{2}}-36b+90 \right)=\left( b-2 \right)\left( 10b-36 \right)$
$\Rightarrow 10{{b}^{2}}-72b+180=10{{b}^{2}}-56b+72\Rightarrow b=\dfrac{27}{4}\Rightarrow {{d}^{2}}\le f\left( \dfrac{27}{4} \right).$
Từ đó $b=\dfrac{27}{4};a=-\dfrac{9}{2};c=-3\Rightarrow a+b+c=-\dfrac{3}{4}.$