1. Given real numbers $x, y$ satisfy $3 x^{2}+4 y^{2}=1$. If $|3 x+4 y-\lambda|+|\lambda+7-3 x-4 y|$ is independent of $x, y$, then the range of values for $\lambda$ is
- 1. $[\sqrt{7}-7,-\sqrt{7}]$.

Let $t=3 x+4 y$. Clearly, when $\lambda \leqslant t \leqslant \lambda+7$, $|t-\lambda|+|t-(\lambda+7)|$ is independent of $x$ and $y$.
By $3 x^{2}+4 y^{2}=1$, let $x=\frac{1}{\sqrt{3}} \cos \alpha, y=\frac{1}{2} \sin \alpha$.
Then $t=3 x+4 y=\sqrt{3} \cos \alpha+2 \sin \alpha$ $=\sqrt{7} \sin (\alpha+\varphi) \in[-\sqrt{7}, \sqrt{7}]$.
Therefore, $\left\{\begin{array}{l}\lambda \leqslant-\sqrt{7}, \\ \lambda+7 \geqslant \sqrt{7}\end{array} \Rightarrow \sqrt{7}-7 \leqslant \lambda \leqslant-\sqrt{7}\right.$.