3. Let the real constant $k$ be such that the equation $2 x^{2}+2 y^{2}-5 x y$ $+x+y+k=0$ represents two intersecting lines in the Cartesian coordinate system $x O y$, with the intersection point being $P$. If points $A$ and $B$ lie on these two lines respectively, and $|\boldsymbol{P A}|=|\boldsymbol{P B}|=1$, then $P A \cdot P B=$ $\qquad$
3. $\pm \frac{4}{5}$.

From the problem, we know that the quadratic polynomial in $x$ and $y$
$$
2 x^{2}+2 y^{2}-5 x y+x+y+k=0
$$

can be factored into the product of two linear factors.
Since $2 x^{2}+2 y^{2}-5 x y=(-2 x+y)(-x+2 y)$,

we have
$$
\begin{array}{l}
2 x^{2}+2 y^{2}-5 x y+x+y+k \\
=(-2 x+y+a)(-x+2 y+b),
\end{array}
$$

where $a$ and $b$ are constants to be determined.
By expanding the above expression and comparing the coefficients of corresponding terms, we get
$$
a b=k, -a-2 b=1, b+2 a=1 \text{. }
$$

Solving these equations, we find $a=1, b=-1, k=-1$.
From $\left\{\begin{array}{l}-2 x+y+1=0, \\ -x+2 y-1=0,\end{array}\right.$ we get the slopes of the two lines as
$$
k_{1}=\frac{1}{2}, k_{2}=2 \text{, and the intersection point } P(1,1) \text{. }
$$

Let the angle between the two lines be $\theta$ (where $\theta$ is acute). Then
$$
\begin{array}{l}
\tan \theta=\frac{k_{2}-k_{1}}{1+k_{2} k_{1}}=\frac{3}{4}, \cos \theta=\frac{4}{5} . \\
\text{ Hence } \boldsymbol{P A} \cdot \boldsymbol{P B}=|\boldsymbol{P A}| \cdot|\boldsymbol{P B}| \cos \theta
\end{array}
$$

or $|\boldsymbol{P A}| \cdot|\boldsymbol{P B}| \cos \left(180^{\circ}-\theta\right)$
$$
= \pm|\boldsymbol{P A}| \cdot|\boldsymbol{P B}| \cos \theta= \pm \frac{4}{5} .
$$