I have three points to write the equation of a plane: assume $P_1=(x_1,y_1,z_1),P_2=(x_2,y_2,z_2),P_3=(x_3,y_3,z_3)$. I can also write the equation of this plane.
I want to obtain the coordinate of a point $(P_4)$ in this plane which is at a particular distance of $P_1, P_2$, whereas $P_1,P_2,P_4$ constitute a right angled triangle with each other.
Could anyone help me with this?
As Gerry Myerson mentioned, for any point $P_4$ that lies on a circle where $\overline{P_1P_2}$ forms the diameter of the circle, the triangle $P_1P_2P_4$ will have a right angle at $P_4$. However, if you know the distances between $P_4$ and $P_1,P_2$ each, you will find two solutions which may or may not form a right angled triangle.. Do you have the absolute distances, or maybe a ratio of distances?
By the way, another way to get right angles would be at $P_1$ or $P_2$.
EDIT:
According to your comment, you have the following conditions:
$(P_4-P_2)^2 = d^2$, so $P_4$ has to be in a sphere with radius $d$ around $P_2$, or on a circle on the plane you already found
To get a right angle in $P_2$, you can choose $P_4$ to be on a line through $P_2$ perpendicular to $\overline{P_1P_2}$ ($P_4$ would be on the intersection of the line and the circle/sphere, so there are two solutions). There are more possible solutions (you could get a line through $P_1$ to get a right angle in $P_1$, but you will have no guarantee that the sphere and this line have an intersection, or you could try to find an intersection of the sphere with the circle where $\overline{P_1P_2}$ is the diameter, but also here there's no guarantee you'll find a solution).
Hope this helps!