Ta có: $C(a;a;0), B'(a;0;a), C(a;a;a), D(0;a;a), M(\frac{a}{2};0;a ), N(a;\frac{a}{2};0 )$
$\overrightarrow{A'N}=(a;\frac{a}{2};-a )=\frac{a}{2}(2;1;-2) $
$\overrightarrow{B'D}=(-a;a;-a) =-a(1;-1;1)$
$\Rightarrow [\overrightarrow{A'N},\overrightarrow{B'D} ]=\frac{a^2}{2} (1;4;3)$
a) Phương trình $(\alpha ):(x-\frac{a}{2} )+4y+3(z-a)=0\Leftrightarrow 2x+8y+6z-7a=0$
b) $V_{A'.NB'D}=\frac{1}{6}|[\overrightarrow{A'N},\overrightarrow{A'B'} ].\overrightarrow{A'D} | $
$=\frac{1}{6}|[(a;\frac{a}{2};-a), (a;0;0) ](0;a;-a)|=\frac{a^3}{12} $ (đvtt)
c) Gọi
$\left\{ \begin{array}{l} \alpha_1=\frac{\overrightarrow{NA'} }{NA} =(\frac{-2}{3};-\frac{1}{3};\frac{2}{3} ) \\ \alpha_2=\frac{\overrightarrow{ND} }{ND}=(-\frac{2}{\sqrt{5} };\frac{1}{\sqrt{5} };0 ) \end{array} \right. $
$\Rightarrow \overrightarrow{\alpha }=\overrightarrow{\alpha_1 }+\overrightarrow{\alpha_2 } =\frac{-1}{3\sqrt{5} } (6+2\sqrt{5};\sqrt{5}-3 ;-2\sqrt{5} ) $ là vecto chỉ phương của $NE$
Vậy phương trình tham số $NE:\left\{ \begin{array}{l} x=a+(6+2\sqrt{5} )t\\ y=\frac{a}{2}+(\sqrt{5}-3 )t \\z=-2\sqrt{5}t \end{array} \right. (t\in R)$
d)
$c{\rm{os}}\varphi = \frac{{|\overrightarrow {A'N} .\overrightarrow {B'D} |}}{{A'N.B'D}} = \frac{{|(2;1; - 2)(1; - 1;1)|}}{{\sqrt {4 + 1 + 4} .\sqrt {1 + 1 + 1} }} = \frac{{\sqrt 3 }}{9}$
$d(A'N,B'D):d = \frac{{|{\rm{[}}\overrightarrow {A'N} {\rm{,}}\overrightarrow {B'D} {\rm{]}}{\rm{.}}\overrightarrow {A'B} |}}{{|{\rm{[}}\overrightarrow {A'N} {\rm{,}}\overrightarrow {B'D} {\rm{]}}|}} = \frac{{\left| {\frac{{{a^2}}}{2}(1;4;3)(a;0;0)} \right|}}{{\frac{{{a^2}}}{2}\sqrt {1 + 16 + 9} }} = \frac{a}{{\sqrt {26} }}$