Question

The point which is equidistant from the points $(-1,1,3),(2,1,2),(0,5,6)$ and $(3,2,2)$ is

• A. $(-1,3,4)$
• B. $(3,1,4)$
• C. $(1,3,4)$
• D. $(4,1,3)$

Answer 1

Answer: (C)

$(1,3,4)$

Let $(x,y,z)$ be the points equidistant from the given points by the distance $d$.

Then, we have the equations

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

$(x-2{)}^2+(y-1{)}^2+(z-2{)}^2=d^2$

$x^2+(y-5{)}^2+(z-6{)}^2=d^2$

$(x-3{)}^2+(y-2{)}^2+(z-2{)}^2=d^2$

Using the equation 1 and 2 we get,

$z=3x+1$

And using equations 2 and 4 we get,

$y=4-x$.

Substituting these values in equation 1 and 3 we get,

$x^2+2x+1+9-6x+x^2+9x^2-12x+4$

$=x^2+x^2+2x+1+9x^2-30x+25$

$\Rightarrow x=1$.

Hence, the point is given by $(x,4-x,3x+1)=(1,3,4)$.