3. Given the parabola $y=x^{2}+k x+4-k$ intersects the $x$-axis at integer points $A, B$, and intersects the $y$-axis at point $C$. Then $S_{\triangle A B C}=$ $\qquad$ .
3.24.

Given the equation $x^{2}+k x+4-k=0$ has two distinct integer roots $x_{1}, x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-k, \\
x_{1} x_{2}=4-k .
\end{array}\right.
$$
(2) - (1) gives $x_{1}+x_{2}-x_{1} x_{2}+4=0$,
$$
\left(x_{1}-1\right)\left(x_{2}-1\right)=5 \text {. }
$$

Thus, $\left\{\begin{array}{l}x_{1}-1=1, \\ x_{2}-1=5\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}-1=5, \\ x_{2}-1=1\end{array}\right.$
or $\left\{\begin{array}{l}x_{1}-1=-1, \\ x_{2}-1=-5\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}-1=-5, \\ x_{2}-1=-1 .\end{array}\right.$
which means $x_{1}+x_{2}=8$ or $x_{1}+x_{2}=-4$.
Therefore, $k_{1}=-8$ or $k_{2}=4$.

When $k=-8$, from $x^{2}-8 x+12=0$, we get $x_{1}=2$, $x_{2}=6$. Thus, $A B=4$, and the coordinates of point $C$ are $(0,12)$.
$$
\therefore S_{\triangle A B C}=\frac{1}{2} \times 4 \times 12=24 \text {. }
$$

When $k=4$, from $x^{2}+4 x=0$, we get $x_{1}=0, x_{2}=-4$, then $A B=4$, and $C(0,0)$. In this case, points $A, B, C$ cannot form a triangle. Hence, $S_{\triangle A B C}=24$.