In 3D find a point on plane with given plane angle, distance and three other points I have three points $x, y, z$ in 3D which are not on the same line. With this points we can build a plain $P$. Is there any way to get point $z'$ on $P$ which is at a distance of $d$ from $y$ and angle between $xy$ and $yz'$ is $\alpha$? I'm looking for some algorithm for that.
You could start thinking the particular case when $Z=0$; i.e. when you are in $\mathbb{R}^2$. Put $y$ at the origin and draw all the points with distance $d$ from the origin (a circle with radius $d$ and center at $(0,0)$). Does this gives you a clue?
Say $\vec x, \vec y,\vec z, \vec {z’} $ are the position vectors of the respective points, with $$\vec{z’}=(a,b,c) $$ I’ll show you how you can get $3$ equations in the $3$ unknowns: $a,b,c$.
Firstly, we know $x,y,z’$ are coplanar, and so $$ (a,b,c) \cdot (\vec {xy} \times \vec {yz})=0 \hspace{1 cm} \mathbf{(1)} $$ Also, $|\vec{yz’} | = d$, so $$|(a,b,c)-\vec y | = d \hspace{1 cm} \mathbf{(2)}$$
Lastly, taking into account the angle, $$\cos\alpha = \frac{\vec{xy}\cdot\vec{yz’}}{|\vec{xy}| |\vec{yz’}|} \\ = \frac{(\vec y-\vec x)\cdot (\vec{z’} -\vec y)}{|\vec y-\vec x| d } 
\\ \implies (a,b,c)\cdot (\vec y-\vec x) = d\cos\alpha |\vec y-\vec x| + \vec y \cdot(\vec y-\vec x) \hspace{1 cm} \mathbf{(3)} $$
Just solve these equations to get $a,b,c$ and consequently $\vec{z’}$.