11. The point $P$ on the line $3 x+2 y=1$ is equidistant from points $A(2,1)$ and $B(1,-2)$. Then the coordinates of point $P$ are $\qquad$.
$$
=11 \cdot\left(\frac{3}{7},-\frac{1}{7}\right) \text {. }
$$

Let $P(x, y)$. From $|P A|=|P B|$, we know $x+3 y=0$.
Also, $3 x+2 y=1$, solving the system of equations yields
$$
x=\frac{3}{7}, y=-\frac{1}{7} \text {. }
$$