6. Given real numbers $x, y, z$ satisfy $x+y+z=1$. Then the maximum value of $M=xy+2yz+3xz$ is ( ).
(A) $\frac{1}{2}$
(B) $\frac{2}{3}$
(C) $\frac{3}{4}$
(D) 1
6. C.

From the given information,
$$
\begin{array}{l}
M=x y+2 y z+3 x z \\
=x y+(2 y+3 x)(1-x-y) \\
=-3 x^{2}-4 x y-2 y^{2}+3 x+2 y \\
=-2\left(y+x-\frac{1}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4} \\
\leqslant \frac{3}{4} .
\end{array}
$$

Therefore, the maximum value of $M$ is $\frac{3}{4}$.