Trong không gian với hệ tọa độ $Oxyz$ cho hai điểm $A\left( 1;2;-1...

Cách 1: PP Trắc nghiệm
Gọi $\left( R \right)$ : $\left\{ \begin{aligned}
& A,B\in \left( R \right) \\
& \left( R \right)\bot \left( P \right) \\
\end{aligned} \right.\Rightarrow \overrightarrow{{{n}_{\left( R \right)}}}=\left[ \overrightarrow{AB};\overrightarrow{{{n}_{\left( p \right)}}} \right]=\left( -3;0;-3 \right)=-3\left( 1;0;1 \right).$
Mặt phẳng $\left( Q \right)$ đi qua hai điểm $A$, $B$ và tạo với mặt phẳng $\left( P \right)$ một góc nhỏ nhất
$\Leftrightarrow \left\{ \begin{aligned}
& A,B\in \left( Q \right) \\
& \left( Q \right)\bot \left( R \right) \\
\end{aligned} \right.\Rightarrow \overrightarrow{{{n}_{\left( Q \right)}}}=\left[ \overrightarrow{AB};\overrightarrow{{{n}_{\left( R \right)}}} \right]=\left( 2;2;-2 \right)=2\left( 1;1;-1 \right).$
Chọn $\overrightarrow{{{n}_{\left( Q \right)}}}=\left( 1;1;-1 \right)\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& b=-1 \\
\end{aligned} \right.\Rightarrow a+b=1+-1=0.$
Cách 2: PP Tự luận
$\left( Q \right)$ có một véctơ pháp tuyến là $\overrightarrow{{{n}_{\left( Q \right)}}}=\left( 1;a;b \right)$ và đi qua hai điểm $A\left( 1;2;-1 \right),B\left( 0;4;0 \right)$.
Suy ra: $\overrightarrow{{{n}_{\left( Q \right)}}}.\overrightarrow{AB}=0\Leftrightarrow -1+2a+b=0\Leftrightarrow b=1-2a\Rightarrow \overrightarrow{{{n}_{\left( Q \right)}}}=\left( 1;a;1-2a \right).$
$\cos \widehat{\left( \left( P \right);\left( Q \right) \right)}=\dfrac{\left| \overrightarrow{{{n}_{\left( P \right)}}}.\overrightarrow{{{n}_{\left( Q \right)}}} \right|}{\left| \overrightarrow{{{n}_{\left( P \right)}}} \right|.\left| \overrightarrow{{{n}_{\left( Q \right)}}} \right|}=\dfrac{\left| 2.1-1.a-2\left( 1-2a \right) \right|}{\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}}.\sqrt{{{1}^{2}}+{{a}^{2}}+{{\left( 1-2a \right)}^{2}}}}=\dfrac{\left| a \right|}{\sqrt{5{{a}^{2}}-4a+2}}$.
Nếu $a=0$ thì $\cos \widehat{\left( \left( P \right);\left( Q \right) \right)}=0\Rightarrow \widehat{\left( \left( P \right);\left( Q \right) \right)}={{90}^{0}}\Rightarrow {{\widehat{\left( \left( P \right);\left( Q \right) \right)}}_{\max }}(l).$
Nếu $a\ne 0$ thì
$\cos \widehat{\left( \left( P \right);\left( Q \right) \right)}=\dfrac{\left| a \right|}{\sqrt{5{{a}^{2}}-4a+2}}=\dfrac{1}{\sqrt{5-4.\dfrac{1}{a}+2.{{\left( \dfrac{1}{a} \right)}^{2}}}}=\dfrac{1}{\sqrt{2.{{\left( \dfrac{1}{a}-1 \right)}^{2}}+3}}\le \dfrac{1}{\sqrt{3}}$.
Suy ra
$\begin{aligned}
& {{\widehat{\left( \left( P \right);\left( Q \right) \right)}}_{\min }}\Leftrightarrow \cos {{\widehat{\left( \left( P \right);\left( Q \right) \right)}}_{\max }} \\
& \Leftrightarrow \cos \widehat{\left( \left( P \right);\left( Q \right) \right)}=\dfrac{1}{\sqrt{3}}\Leftrightarrow \dfrac{1}{a}=1\Leftrightarrow a=1\Rightarrow \overrightarrow{{{n}_{\left( Q \right)}}}=\left( 1;1;-1 \right). \\
\end{aligned}$
$\overrightarrow{{{n}_{\left( Q \right)}}}=\left( 1;1;-1 \right)\Rightarrow \left\{ \begin{aligned}
& a=1 \\
& b=-1 \\
\end{aligned} \right.\Rightarrow a+b=1+-1=0.$