11. Let $F$ be the focus of the parabola $y^{2}=2 x-1$, and $Q(a, 2)$ be a point on the line $y=2$. If there is exactly one point $P$ on the parabola such that $|P F|=|P Q|$, then the value of $a$ is $\qquad$
11.0 or 1.

It is known that the perpendicular bisector of $F(1,0), Q F$ is
$$
y=\frac{1-a}{2} x+\frac{a^{2}+3}{4} \text {. }
$$

It intersects the parabola at only 1 point.
$$
\left\{\begin{array}{l}
y=\frac{1-a}{2} x+\frac{a^{2}+3}{4}, \\
y^{2}=2 x-1 .
\end{array}\right.
$$

When $a=1$, it is obvious that there is only 1 intersection point.
When $a \neq 1$, substituting equation (1) into equation (2) yields
$(1-a) y^{2}=\left(y-\frac{a^{2}+3}{4}\right) \times 4-(1-a)$.
Rearranging gives $(1-a) y^{2}-4 y+a^{2}-a+4=0$.
By $\Delta=16-4(1-a)\left(a^{2}-a+4\right)=0$
$\Rightarrow a^{3}-2 a^{2}+5 a=0$
$\Rightarrow a\left(a^{2}-2 a+5\right)=0 \Rightarrow a=0$.