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

Answer 1

The correct option is C $(1,3,4)$

Given: $A(-1,1,3),B(2,1,2),C(0,5,6)$ and $D(3,2,2)$
Let $(x,y,z)$ be the points equidistant from the given points by the distance $d$.
Then, we have the equations
Using Distance formula,
$(x+1{)}^2+(y-1{)}^2+(z-3{)}^2=d^2\cdots (i)$
$(x-2{)}^2+(y-1{)}^2+(z-2{)}^2=d^2\cdots (ii)$
$(x-0{)}^2+(y-5{)}^2+(z-6{)}^2=d^2\cdots (iii)$
$(x-3{)}^2+(y-2{)}^2+(z-2{)}^2=d^2\cdots (iv)$
Using equation $\left(i\right),\left(ii\right)$ we get,
$z=3x+1$
Using equation $\left(ii\right),\left(iv\right)$ we get,
$y=4-x$
Putting the values in equation$\left(i\right),\left(iii\right)$
$\Rightarrow x^2+2x+1+9-6x+x^2+9x^2-12x+4=x^2+x^2+2x+1+9x^2-30x+25$
$\Rightarrow x=1$
Hence, the point is given by $(x,4-x,3x+1)=(1,3,4)$