The circumcentre of the triangle formed by the points $(1,2,3),(3,-1,5),(4,0,-3)$ is

Question

Aicharya · Accepted Answer

Let $A, B, C$ be the vertices of the triangle given as, $\mathrm{A}(1,2,3), \mathrm{B}(3,-1,5), \mathrm{C}(4,0,-3)$

Now, let $O(x, y, z)$ be the circumcentre of $\triangle A B C$. $ \begin{aligned} & \therefore \quad O A=O B=O C \\ & O A=O B \Rightarrow O A^2=O B^2 \\ & \begin{aligned} \Rightarrow \quad(x-1)^2+(y-2)^2+(z-3)^2 \\ \quad=(x-3)^2+(y+1)^2+(z-5)^2 \end{aligned} \\ & x^2-2 x+1+y^2-4 y+4+z^2-6 z+9 \\ & =x^2-6 x+9+y^2+2 y+1+z^2-10 z+25 \end{aligned} $

Similarly, $O B=O C$ $ \begin{aligned} & (x-3)^2+(y+1)^2+(z-5)^2 \\ & =(x-4)^2+(y-0)^2+(z+3)^2 \\ & x^2-6 x+9+y^2+2 y+1+z^2-10 z+25 \\ & =x^2-8 x+16+y^2+z^2+6 z+9 \end{aligned} $

Similarly, $O A=O C$ $ \begin{gathered} (x-1)^2+(y-2)^2+(z-3)^2 \\ =(x-4)^2+(y-0)^2+(z+3)^2 \\ x^2-2 x+1+y^2-4 y+4+z^2-6 z+9 \\ =x^2-8 x+16+y^2+z^2+6 z+9 \\ \Rightarrow \quad 6 x-4 y-12 z-11=0 \end{gathered} $ On solving Eqs. (i), (ii) and (iii), we get $x=\frac{7}{2}, y=-\frac{1}{2}, z=1$ $\therefore \quad$ Circumcentre of $\triangle A B C$ is $\left(\frac{7}{2}, \frac{-1}{2}, 1\right)$

The circumcentre of the triangle formed by the points $(1,2,3),(3,-1,5),(4,0,-3)$ is

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