6. Given two sets of points on a plane
$$
\begin{aligned}
M= & \left\{(x, y)|| x+y+1 \mid \geqslant \sqrt{2\left(x^{2}+y^{2}\right)},\right. \\
& x, y \in \mathbf{R}\}, \\
N= & \{(x, y)|| x-a|+| y-1 \mid \leqslant 1, x 、 y \in \mathbf{R}\} .
\end{aligned}
$$

If $M \cap N \neq \varnothing$, then the range of values for $a$ is
$\qquad$
6. $[1-\sqrt{6}, 3+\sqrt{10}]$.

From the problem, $M$ is the set of points on a parabola with the origin as the focus and the line $x+y+1=0$ as the directrix, and its concave side. $N$ is the set of points in a square and its interior with center $(a, 1)$ (as shown in Figure 2).

Consider the case when $M \cap N=\varnothing$, the range of $a$:
Let $y=1$, substitute into the equation $|x+y+1|=\sqrt{2\left(x^{2}+y^{2}\right)}$, we get $x^{2}-4 x-2=0$. Solving this, we get $x=2 \pm \sqrt{6}$.
Therefore, when $a3+\sqrt{10}$, $M \cap N=\varnothing$.
Thus, when $1-\sqrt{6} \leqslant a \leqslant 3+\sqrt{10}$, i.e., $a \in[1-\sqrt{6}$, $3+\sqrt{10}]$, $M \cap N=\varnothing$.