Phương trình mặt phẳng - Lý thuyết

Ta có:$\left( P \right):\,\,\,2x - 2y + z - 2 = 0$

$\Rightarrow d\left( {M;\,\,\left( P \right)} \right) = \dfrac{{\left| {2.1 - 2.6 - 3 - 2} \right|}}{{\sqrt {{2^2} + {{\left( { - 2} \right)}^2} + 1} }}$ $= \dfrac{{15}}{3} = 5.$

$d\left( {\left( P \right),\left( Q \right)} \right) = \dfrac{{\left| { - 11 - 4} \right|}}{{\sqrt {{2^2} + {2^2} + {{\left( { - 1} \right)}^2}} }} = 5.$

Đặt $\left( \alpha  \right):\ 2x+y+mz-1=0.$

Ta có: $d\left( A;\ \left( \alpha  \right) \right)=\dfrac{\left| 2.1+2+3.m-1 \right|}{\sqrt{{{2}^{2}}+{{1}^{2}}+{{m}^{2}}}}=\dfrac{\left| 3+3m \right|}{\sqrt{{{m}^{2}}+5}}.$

$\begin{array}{l}\overrightarrow {AB}  = \left( {2;\;2;\;1} \right) \Rightarrow AB = \sqrt {{2^2} + {2^2} + 1}  = 3.\\ \Rightarrow d\left( {A;\left( \alpha  \right)} \right) = AB \Leftrightarrow \dfrac{{\left| {3 + 3m} \right|}}{{\sqrt {{m^2} + 5} }} = 3\\ \Leftrightarrow \left| {m + 1} \right| = \sqrt {{m^2} + 5} \\ \Leftrightarrow {m^2} + 2m + 1 = {m^2} + 5\\ \Leftrightarrow m = 2.\end{array}$