3. Let the chord $P Q$ of the parabola $y^{2}=x$ be perpendicularly bisected by the line $l: y=k(x-1)+1(k \in \mathbf{Z})$. Then the length of the chord $P Q$ is $\qquad$ .
3. $\sqrt{10}$.

Let the equation of line $P Q$ be $y=-\frac{1}{k} x+b$ (obviously $k \neq 0$, otherwise, $l$ cannot perpendicularly bisect $P Q$).
From $\left\{\begin{array}{l}y^{2}=x, \\ y=-\frac{1}{k} x+b,\end{array}\right.$ eliminating $x$ and rearranging gives
$$
y^{2}+k y-b k=0 \text {. }
$$

Since $P Q$ intersects the parabola $y^{2}=x$ at two distinct points, the discriminant of the above equation is greater than zero, i.e.,
$$
k^{2}+4 b k>0 \text {. }
$$

Let the midpoint of $P Q$ be $M$, then
$$
y_{M}=-\frac{k}{2}, x_{M}=\frac{k^{2}}{2}+b k \text {. }
$$

Since $M$ lies on line $l$, we have
$$
-\frac{1}{2} k=k\left(\frac{1}{2} k^{2}+b k-1\right)+1 \text {. }
$$

Substituting equation (2) into equation (1) and rearranging gives
$$
\frac{1}{k}(k+2)\left(k^{2}-2 k+2\right)<0 \text {. }
$$

Solving this, we get $-2<k<0$.
Since $k \in \mathbf{Z}$, we have $k=-1$.
Substituting $k=-1$ into equation (2), we get $b=-1$.
Thus, the equation of line $P Q$ is $y=x-1$.
From $\left\{\begin{array}{l}y=x-1, \\ y^{2}=x\end{array}\right.$ eliminating $y$, we get $x^{2}-3 x+1=0$.
Let $x_{1}$ and $x_{2}$ be its two roots, by Vieta's formulas we have
$$
\begin{array}{l}
x_{1}+x_{2}=3, x_{1} x_{2}=1 . \\
\text { Therefore }|P Q|=\sqrt{1+k_{P Q}^{2}}\left|x_{1}-x_{2}\right| \\
=\sqrt{2} \cdot \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\sqrt{10} .
\end{array}
$$