I'm interested in visualising the following surface of points $(x,y,z)\in \mathbb{R}^3$ which satisfy the following equation and inequalities:
$$ \begin{cases} x + y + z = 0\\ x \leq y \leq z \end{cases} $$
The first equation determines a plane passing through the origin $(0,0,0)$, therefore the inequalities (which I have written in compact form) should determine a section of this plane. Another way would be to start from the part of space determined by $x<y<z$ and then see the intersection with the plane, but I can't manage to visualize it and I can't seem to get it to appear on 3D online graphing tools.
One thing that I thought might help is to project it to the $yz$ plane with the substitution $x = -y-z$, where the surface reduces to an angle delimited by $z=y$ and $z=-2y$:
$$ \begin{cases} z + 2y \geq 0\\ z - y \leq 0 \end{cases} $$
However, now I don't know how to incorporate this figure in 3D with the additional constraint $x=-y-z$. Is it like a "slice" of 3D space with a slant?
Beyond this, I am also interested in the more general $n$-dimensional problem with $(x_1,\ldots,x_n)\in\mathbb{R}^n$ such that:
$$ \begin{cases} x_1 + \ldots + x_n = 0\\ x_1 \leq x_2 \leq \ldots \leq x_n \end{cases} $$
So some intuition for these higher dimensional cases would also be useful. Any help?
Establish an orthonormal coordinate system on the plane $x + y + z = 0$. Two orthonormal vectors on it is $\frac1{\sqrt 2}(0, -1, 1)$ and $\frac 1{\sqrt 6}(-2,1,1)$. An arbitrary point $(x,y,z)$ on the plane should be some linear combination $au + bv$. Or $$x = -\frac2{\sqrt6}b\\y =-\frac a{\sqrt 2} +\frac b{\sqrt 6}\\z =\frac a{\sqrt 2} + \frac b{\sqrt 6}$$ The inequalities then become $$y \ge x \iff -\frac a{\sqrt 2} +\frac b{\sqrt 6} \ge -\frac2{\sqrt6}b \iff b \ge \frac a{\sqrt 3}\\ z \ge y \iff \frac a{\sqrt 2} + \frac b{\sqrt 6} \ge -\frac a{\sqrt 2} +\frac b{\sqrt 6} \iff a \ge 0$$
Which is the region between the positive $b$-axis, and the positive ray of the line $b = \frac a{\sqrt 3}$, which is at an angle of $60^\circ$ with the positive $b$-axis. So your region is a $60^\circ$ angle in the plane and its interior.