Question

Find the point in yz-plane which is equidistant from the points A(3, 2, -1), B(1, -1, 0) and C(2, 1, 2).

Answer 1

Any point in the yz-plane is of the form P(0, y, z).

$|AP|=|BP|=|CP|\iff A{P}^2=B{P}^{2}andB{P}^2=C{P}^2$.

$(0-3{)}^2+(y-2{)}^2+(z+1{)}^2=(0-1{)}^2+(y+1{)}^2+(z-0{)}^2$

and $(0-1{)}^2+(y+1{)}^2+(z-0{)}^2=(0-2{)}^2+(y-1{)}^2+(z-2{)}^2$

$\Rightarrow 3y-z-6=0and4y+4z-7=0$.

On solving, we get $y=\frac{31}{16}$ and $z=\frac{-3}{16}$

Hence, the required point is $(0,\frac{31}{16},\frac{-3}{16})$.