6. $\{(1,2,3,4),(1,2,3,-4),(-1,-2$, $-3,4),(-1,-2,-3,-4)\}$.
(1) - (2) gives
$$
x^{2}+y^{2}+z^{2}=14 \text {. }
$$
(1) $+2 \times$ (2) gives
$$
x^{2}+4 y^{2}+4 z^{2}+3 w^{2}=101 \text {. }
$$
(3) + (5) gives
$$
x^{2}+4 y^{2}+9 z^{2}+4 x y+12 y z+6 z x=196 \text {. }
$$

Thus, $(x+2 y+3 z)^{2}=196$
$$
\Rightarrow x+2 y+3 z= \pm 14 \text {. }
$$

Therefore, the original system of equations is equivalent to
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=14, \\
y^{2}+z^{2}+w^{2}=29 \\
x+2 y+3 z=14
\end{array}\right.
$$

or $\left\{\begin{array}{l}x^{2}+y^{2}+z^{2}=14, \\ y^{2}+z^{2}+w^{2}=29, \\ x+2 y+3 z=-14 .\end{array}\right.$
If the system of equations (6) has a solution, then the line $x+2 y=14-3 z$ intersects the circle $x^{2}+y^{2}=14-z^{2}$, which requires and is only satisfied if
$$
\begin{array}{l}
\frac{|3 z-14|}{\sqrt{1^{2}+2^{2}}} \leqslant \sqrt{14-z^{2}} \\
\Leftrightarrow z^{2}-6 z+9 \leqslant 0 \Rightarrow z=3 .
\end{array}
$$

Similarly, $x=1, y=2$.
Substituting into equation (2) gives $w= \pm 4$.
If the system of equations (7) has a solution, then the line $x+2 y=-14-3 z$ intersects the circle $x^{2}+y^{2}=14-z^{2}$, which requires and is only satisfied if
$$
\begin{array}{l}
\frac{|-3 z-14|}{\sqrt{1^{2}+2^{2}}} \leqslant \sqrt{14-z^{2}} \\
\Leftrightarrow z^{2}+6 z+9 \leqslant 0 \Rightarrow z=-3 .
\end{array}
$$

Similarly, $x=-1, y=-2$.
Substituting into equation (2) gives $w= \pm 4$.
Therefore, the solution set of the original system of equations is
$$
\begin{array}{l}
(x, y, z, w) \\
=\{(1,2,3,4),(1,2,3,-4), \\
\quad(-1,-2,-3,4),(-1,-2,-3,-4)\} .
\end{array}
$$
6. The solution set of the system of equations in $x, y, z, w$
$$
\left\{\begin{array}{l}
x^{2}+2 y^{2}+2 z^{2}+w^{2}=43, \\
y^{2}+z^{2}+w^{2}=29, \\
5 z^{2}-3 w^{2}+4 x y+12 y z+6 z x=95
\end{array}\right.
$$

is $\qquad$ .