11 Given $\alpha, \beta \in \mathbf{R}$, the intersection point of the lines $\frac{x}{\sin \alpha+\sin \beta}+\frac{y}{\sin \alpha+\cos \beta}=1$ and $\frac{x}{\cos \alpha+\sin \beta}+$ $\frac{y}{\cos \alpha+\cos \beta}=1$ lies on the line $y=-x$, then $\sin \alpha+\cos \alpha+\sin \beta+$ $\cos \beta=$ $\qquad$
11 Given that the intersection point of the two lines can be set as $\left(x_{0},-x_{0}\right)$, and $\sin \alpha, \cos \alpha$ are the roots of the equation
$$
\frac{x_{0}}{t+\sin \beta}+\frac{-x_{0}}{t+\cos \beta}=1
$$

which can be rewritten as
$$
t^{2}+(\cos \beta+\sin \beta) t+\sin \beta \cos \beta-x_{0}(\cos \beta-\sin \beta)=0
$$

Therefore,
$$
\sin \alpha+\cos \alpha=-(\sin \beta+\cos \beta),
$$

which means $\sin \alpha+\cos \alpha+\sin \beta+\cos \beta=0$.