6. In the right triangle $\triangle A B C$, the right-angle sides $A C=3$ and $B C=4$. Line $l$ intersects the two right-angle sides $A C$ and $B C$ at points $S$ and $T$ respectively, and $S$ and $T$ are on the interiors of $A C$ and $B C$, and $l$ bisects the perimeter and area of $\triangle A B C$. Then ().
(A) There is only 1 such line
(B) There are only 2 such lines
(C) There are more than 2 such lines
(D) Such a line does not exist
6. (D).

As shown in Figure 4, it is easy to know that the perimeter of $\triangle ABC$ is 12, and the area is 6. If line $l$ bisects the perimeter and area of $\triangle ABC$, then
$$
CS + CT = 6, S_{\triangle CST} = 3.
$$

Let $CS = x$, then $CT = 6 - x$. Thus, $\frac{1}{2} x(6 - x) = 3$.
Solving for $x$ gives $x = 3 \pm \sqrt{3}$.
Since $CS > 4$, i.e., $T$ is outside of $BC$, this leads to a contradiction. Therefore, such a line $l$ does not exist.