If your triangle is in space, the given information doesn't determine it yet, because given any such triangle, rotating it around the line joining $p1$ and $p2$ gives you another valid triangle. If your triangle is in the plane, then the information does determine p3 as long as you decide whether the order $p1$, $p2$, $p3$ is clockwise or counterclockwise. Let's say then, the triangle is in the plane and, as shown in your neat ASCII picture, $p1$, $p2$, $p3$ is clockwise.
If you know about complex numbers, then $p3-p1 = -iB/A(p2-p1)$ (because multiplication by $B/A$ changes a length $A$ vector into a length $B$ vector, and multiplication by $-i$ rotates by $90^\circ$ clockwise).
If you don't want to use complex numbers, then say $p1=(x1, y1)$, $p2=(x2, y2)$ and $p3=(x3, y3)$. Since $p1p2p3$ is a right angle, the slopes of $p1p2$ and $p1p3$ multiply to $-1$, that gives you one equation for $x3$ and $y3$. Additionally, you know that the distance from $p1$ to $p3$ is B, which gives you a second equation for $x3$ and $y3$. The system formed by those equations will have two solutions, one corresponding to $p1$, $p2$, $p3$ being clockwise and the other corresponding to counterclockwise.