6. Let $n$ be a positive integer. $0<x \leqslant 1$. In $\triangle A B C$, if $A B=n+x, B C=n+2 x, C A=n+3 x, B C$ has a height $A D=n$. Then, the number of such triangles is ( ).
(A) 10
(B) 11
(C) 12
(D) infinitely many
6. (C).

Let $\triangle A B C$ be the condition; then $c=n+1, a=n+2 x$, $b=n+3 x, p=\frac{1}{2}(a+b+c)=\frac{1}{2}(3 n+6 x)$. By Heron's formula, we get
$$
\begin{array}{l}
\sqrt{\frac{3 n+6 x}{2} \cdot \frac{n+4 x}{2} \cdot \frac{n+2 x}{2} \cdot \frac{n}{2}}=\frac{n(n+2 x)}{2} . \\
\therefore \frac{3 n(n+2 x)^{2}(n+4 x)}{16}=\frac{n^{2}(n+2 x)^{2}}{4},
\end{array}
$$

which simplifies to
$$
\begin{array}{l}
\frac{3(n+4 x)}{4}=n . \\
\therefore n=12 x . \\
\because 0<x \leqslant 1, n \text { is a positive integer. } \\
\therefore(x, n)=\left(\frac{1}{12}, 1\right),\left(\frac{2}{12}, 2\right), \cdots,\left(\frac{12}{12}, 12\right) .
\end{array}
$$

Therefore, there are 12 such triangles.