Get rectangular coordinates of a 3d point, with the polar coordinates? Ok, so imagine a cannon on an cartesian space. X and Y are sideways and Z is upwards. It can rotate horizontally on the world's XY axis (right is clockwise, and left is counter-clockwise), and also vertically (on the cannon's local YZ axis. down is clockwise and up is counterclocwise).
I like to call the angles "θ" for rotation, and "δ" for vertical
Now the cannon shoots something that travels for an N distance.

Where is the particle now, in (x, y, z) coordinates?

In 2d, it's easy, we just need to use: x = length * cos(angle)
y = length * sin(angle).
I looked into Spherical Coordinates. But I tested in a calculator, and if you had a point at (X; 45°; 45°), in the system i'm using, the line from it to the origin should be the diagonal of a cube with side $ X / √3$. But while it gives me an equal X and Y, the Z is different.
 A: I think you need spherical coordinates. Here is the conversion from and to in your case



*

*Spehrical $(r, \delta, \theta)$ to cartesian $(x,y,z)$
$$ \pmatrix{x \\ y \\ z} = \pmatrix{ r \cos \delta \cos \theta \\ r \cos\delta \sin \theta \\ r \sin \delta } $$

*Cartesian $(x,y,z)$ to spherical $(r, \delta, \theta)$
$$ \begin{aligned}
  r & = \sqrt{x^2+y^2+z^2} \\
  \delta & = \mathrm{atan}\left(\frac{z}{\sqrt{x^2+y^2}}\right) \\
  \theta & = \mathrm{atan} \left( \frac{y}{x} \right)
 \end{aligned} $$
For the specific case where $\delta = \theta = 45°$ you get a line along the diagonal
$$ \pmatrix{ x = \frac{1}{2} r \\ y = \frac{1}{2} r \\ z = \frac{1}{\sqrt{2}} r } $$
A: Angles do not occur in isolation. They are measurements to some reference, and that reference is missing from your post, so I am going to make some guesses. (Also "right", "left", "clockwise" and "counter-clockwise" are meaningless without some context on the direction one is looking.)
I'll assume that $\theta$ is the angle between the $xz$ plane (that is, the plane $y = 0$) and the vertical plane of the cannon (that is, the plane containing the $z$-axis, and the "cannon axis" = the central or aiming axis of the cannon), with $\theta$ being positive for counter-clockwise rotations, as seen from above. $\theta$ is often called the "heading" or "yaw" angle. While $\delta$ is the angle of elevation, or "pitch" angle, which is the angle the cannon axis makes with the horizontal plane $z = 0$. That is, the angle the cannon axis makes with its projection in the horizontal plane. $\delta$ is positive when the cannon vector points above the plane, and negative when it points below.
$\theta, \delta = 0$ gives $(1, 0, 0)$ as the default aiming vector. First we rotate it by $\theta$ to get $\vec p = (\cos \theta, \sin \theta, 0)$. Then rotate $\vec p$ up towards the positive $z$-axis (whose direction is the vector $\vec k = (0,0,1)$). The result is given by $\vec c = (\cos \delta)\vec p + (\sin \delta)\vec k$. The result is $$\vec c = (\cos \delta\cos \theta, \cos \delta\sin \theta, \sin \delta)$$ which is, of course, one form of spherical coordinates (the other form being measuring $\phi$ from the positive $z$-axis, which results in $\cos \phi = \sin \delta$ and $\sin \phi = \cos \delta$).
Now as you've noticed, when $\theta = \delta = 45^\circ$, we have $$\vec c = \left(\frac{\sqrt 2}2\frac{\sqrt 2}2, \frac{\sqrt 2}2\frac{\sqrt 2}2, \frac{\sqrt 2}2\right) = \left(\frac 12, \frac 12, \frac{\sqrt 2}2\right)$$
not the point $\vec q = \left(\frac{\sqrt 3}3, \frac{\sqrt 3}3, \frac{\sqrt 3}3\right)$ you were expecting.
Why not? Because that isn't how you defined your angles. The projection of $\vec q$ into the horizontal plane is just $\left(\frac{\sqrt 3}3, \frac{\sqrt 3}3, 0\right)$, which when normalized (i.e., divided by its length to make it a unit vector) is $$\vec q_h = \left(\frac{\sqrt 2}2, \frac{\sqrt 2}2, 0\right)$$
The yaw angle for $\vec q$ is the angle between $\vec q_h$ and $\vec i = (1,0,0)$, which is $\cos^{-1}(\vec q_h \cdot \vec i) = \cos^{-1}\left(\frac{\sqrt 2}2\right) = 45^\circ$ (since $\|\vec q_h\| = \|\vec i\| = 1$). But the pitch angle for $\vec q$ is the angle between $\vec q$ and $\vec q_h$, which is
$$\begin{align}\delta_q &= \cos^{-1}(\vec q_h \cdot \vec q)\\&= \cos^{-1}\left(\frac{\sqrt 3}3\frac{\sqrt 2}2 + \frac{\sqrt 3}3\frac{\sqrt 2}2  + \frac{\sqrt 3}3\,0\right)\\&= \cos^{-1}\left(\frac{\sqrt 6}3\right)\\&\approx 35.26^\circ\end{align}$$
So your assumption that the angle of elevation for $(1,1,1)$ would be $45^\circ$ was mistaken. And when you think it out, it makes sense. $45^\circ$ is equidistant between two axes in a plane, but when you move it over to the location equidistant to all three axes, you move it out of that shortest-distance-plane, which means that it has to move further away from the two axes to get closer to the third, so the angles from the axes has to increase from $45^\circ$. And in fact, the angle between $(1,1,1)$ and each of the coordinate axes is approximately $90^\circ - 35.26^\circ = 54.74^\circ$.
Spherical coordinates (at least the angle of elevation version) is the way you want to go here. The formula you are looking for is $$(N\cos \delta\cos \theta, N\cos \delta\sin \theta, N\sin \delta)$$
You just need to understand a little better how they work.
