6. The set $M=\{u \mid u=12 m+8 n+4 l$, where $m, n, l \in \mathbf{Z}\}$, $N=\{u \mid u=20 p+16 q+12 r$, where $p, q, r \in \mathbf{Z}\}$ has the relationship
A. $M=N$
B. $M \nsubseteq N, N \nsubseteq M$
C. $M \varsubsetneqq N$
D. $M \supsetneqq N$
Notice that $12 m+8 n+4 l=20\left(a_{1} m+b_{1} n+c_{1} l\right)+16\left(a_{2} m+b_{2} n+c_{2} l\right)+12\left(a_{3} m+b_{3} n+c_{3} l\right)$
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array} { l } 
{ 2 0 a _ { 1 } + 1 6 a _ { 2 } + 1 2 a _ { 3 } = 1 2 , } \\
{ 2 0 b _ { 1 } + 1 6 b _ { 2 } + 1 2 b _ { 3 } = 8 , } \\
{ 2 0 c _ { 1 } + 1 6 c _ { 2 } + 1 2 c _ { 3 } = 4 }
\end{array} \Rightarrow \left\{\begin{array}{l}
5 a_{1}+4 a_{2}+3 a_{3}=3, \\
5 b_{1}+4 b_{2}+3 b_{3}=2, \\
5 c_{1}+4 c_{2}+3 c_{3}=1,
\end{array}\right.\right. \\
\text { Take }\left\{\begin{array}{l}
\left(a_{1}, a_{2}, a_{3}\right)=(0,0,1), \\
\left(b_{1}, b_{2}, b_{3}\right)=(1,0,-1), \\
\left(c_{1}, c_{2}, c_{3}\right)=(1,-1,0)
\end{array} \Rightarrow 12 m+8 n+4 l=20(n+l)+16(-l)+12(m-n) .\right.
\end{array}
$$

Similarly, let $20 p+16 q+12 r=12\left(x_{1} p+y_{1} q+z_{1} r\right)+8\left(x_{2} p+y_{2} q+z_{2} r\right)+4\left(x_{3} p+y_{3} q+z_{3} r\right)$
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array} { l } 
{ 3 x _ { 1 } + 2 x _ { 2 } + x _ { 3 } = 5 , } \\
{ 3 y _ { 1 } + 2 y _ { 2 } + y _ { 3 } = 4 , } \\
{ 3 z _ { 1 } + 2 z _ { 2 } + z _ { 3 } = 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
\left(x_{1}, x_{2}, x_{3}\right)=(1,1,0), \\
\left(y_{1}, y_{2}, y_{3}\right)=(0,2,0), \\
\left(z_{1}, z_{2}, z_{3}\right)=(1,0,0)
\end{array}\right.\right. \\
\Rightarrow 20 p+16 q+12 r=12(p+r)+8(p+2 q)+4 \cdot 0 .
\end{array}
$$

Thus $M \subseteq N, N \subseteq M \Rightarrow M=N$. Therefore, the answer is $A$.