Third vertices of isosceles triangle

Question

Find the coordinates of R if you know two vertices of the isosceles triangle having coordinates as P(2,3,4) and Q (4,5,6) and distance of two the vertices PR and QR is 5

Answer 1

PR=QR =5, PQ = sqr(4 +1 +4) = sqr9 = 3

It's a 5-3-3 triangle

P=(2,3,4) PQ = sqr((xr -2)^2 +(yr-3)^2 +(zr-4)^2) =sqr25

xr-2 =2, xr=2 or 0

yr -3 = 1, yr =4 or 2

zr-4 =5, zr =9 or -1

R = (2,4,9) or (0,2,-1)

QR = (4,5,6)

PQ = 5 =sqr( (xr-4)^2 + (yr-5)^2 + (zr-6)^2)

xr-2 =2, xr = 4 or 0

yr-3=1 yr= 4 or 2

zr -6= 5 zr= 11 or 1

R=(4,4,11) or (0,2,1) last triple almost fits almost = (0,2,-1)

keep trying retro engineering with other (xr, yr, zr) values until PQ is the same using (2,3,4) and (4,5,6)

don't give up