6. The line $l:(2 m+1) x+(m+1) y=7 m+4$ is intercepted by the circle $\odot A$ with center $A(1,2)$ and radius 3, the shortest chord length is ( ).
(A) $\sqrt{15}$
(B) 4
(C) $\sqrt{17}$
(D) $3 \sqrt{2}$
6. B.

The equation of line $l$ is transformed into
$$
\begin{array}{l}
m(2 x+y-7)+(x+y-4)=0 . \\
\text { Let }\left\{\begin{array} { l } 
{ 2 x + y - 7 = 0 , } \\
{ x + y - 4 = 0 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
x=3, \\
y=1 .
\end{array}\right.\right.
\end{array}
$$

Therefore, line $l$ passes through point $B(3,1)$.
As shown in Figure 3, connect $A B$, and draw chord $C D$ $\perp A B$ through point $B$. Then $C D$ is the shortest chord intercepted by line $l$, and connect $A C$.
In the right triangle $\triangle A B C$,
$A C=3$,
$A B=\sqrt{5}$.
By the Pythagorean theorem,
$$
B C=\sqrt{A C^{2}-A B^{2}}=2 \text {. }
$$

By the perpendicular chord theorem,
$$
C D=2 B C=4 \text {. }
$$