3. Let the line $y=-\frac{1}{2} x+1$ intersect the $x$-axis and $y$-axis at points $B$ and $A$, respectively. Point $C$ is the reflection of point $B$ over the $y$-axis. Construct an isosceles right triangle $\triangle ACD$ with $AC$ as one of the legs in the second quadrant. Draw $DE \perp x$-axis at point $E$. If the line $y=kx-2k$ divides the quadrilateral $OADE$ into two regions of equal area, then $k=$ . $\qquad$
3. $-\frac{3}{14}$.

As shown in Figure 7, from the given conditions we have
$$
A(0,1), B(2,0), C(-2,0) \text {. }
$$

It is easy to prove that Rt $\triangle D E C \cong$ Rt $\triangle C O A$.
Thus, $D E=C O=2, E C=O A=1, O E=3$.
It is evident that the line $y=k x-2 k$ passes through point $B$, and this line intersects $O A$ and $D E$ at points $M(-2 k, 0)$ and $N(-5 k, 0)$, respectively.
Since $O M+E N=\frac{1}{2}(O A+D E)=\frac{3}{2}$, we have
$$
-2 k-5 k=\frac{3}{2} \Rightarrow k=-\frac{3}{14} .
$$